Quadratic Formula

What Is a Quadratic Equation?

A quadratic equation is any equation that can be written in the form:

$$ax^2 + bx + c = 0$$

where $a$, $b$, and $c$ are numbers (called coefficients) and $a \neq 0$. The key characteristic is that the highest exponent on $x$ is 2. That's what makes it "quadratic" (from the Latin word for "square").

Here are some examples of quadratic equations:

$x^2 - 5x + 6 = 0$
$2x^2 + 7x - 4 = 0$
$x^2 - 16 = 0$
$3x^2 + 12x = 0$

Notice that not all three terms have to be present. You might have a missing $b$ term (like $x^2 - 16 = 0$) or a missing $c$ term (like $3x^2 + 12x = 0$). But you always need the $x^2$ term, because that's what makes it quadratic.

There are several ways to solve quadratic equations: factoring, completing the square, graphing, or using the quadratic formula. The quadratic formula is the most powerful because it works for every quadratic equation, even when factoring is difficult or impossible.

The Quadratic Formula

Here's the formula that solves any quadratic equation:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

This formula gives you the solutions (also called roots or zeros) of the equation $ax^2 + bx + c = 0$. The $\pm$ symbol means "plus or minus," so you'll typically get two solutions: one using addition and one using subtraction.

Let me break down what each part means:

$a$, $b$, and $c$ come from your original equation $ax^2 + bx + c = 0$
The $-b$ on top means you take the opposite of $b$
The $\sqrt{b^2 - 4ac}$ is the square root of $b^2 - 4ac$ (we'll talk more about this part later)
You divide everything by $2a$

The formula might look intimidating, but once you practice it a few times, it becomes routine. Just identify $a$, $b$, and $c$, plug them into the formula, and simplify.

Using the Formula

The key to using the quadratic formula successfully is correctly identifying $a$, $b$, and $c$. Your equation must be in the form $ax^2 + bx + c = 0$ first. If it's not, rearrange it.

Example 1: Solve $x^2 + 5x + 6 = 0$.

First, identify the coefficients:

$a = 1$ (the coefficient of $x^2$)
$b = 5$ (the coefficient of $x$)
$c = 6$ (the constant term)

Now plug into the formula: $$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}$$

Simplify inside the square root: $$x = \frac{-5 \pm \sqrt{25 - 24}}{2}$$ $$x = \frac{-5 \pm \sqrt{1}}{2}$$ $$x = \frac{-5 \pm 1}{2}$$

Now split into two solutions: $$x = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$$ $$x = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$$

The solutions are $x = -2$ and $x = -3$.

Example 2: Solve $2x^2 - 7x + 3 = 0$.

Identify coefficients:

$a = 2$
$b = -7$
$c = 3$

Plug into the formula: $$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$$ $$x = \frac{7 \pm \sqrt{49 - 24}}{4}$$ $$x = \frac{7 \pm \sqrt{25}}{4}$$ $$x = \frac{7 \pm 5}{4}$$

Two solutions: $$x = \frac{7 + 5}{4} = \frac{12}{4} = 3$$ $$x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}$$

The solutions are $x = 3$ and $x = \frac{1}{2}$.

Example 3: Solve $x^2 - 4x - 5 = 0$.

Coefficients:

$a = 1$
$b = -4$
$c = -5$

Apply the formula: $$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}$$ $$x = \frac{4 \pm \sqrt{16 + 20}}{2}$$ $$x = \frac{4 \pm \sqrt{36}}{2}$$ $$x = \frac{4 \pm 6}{2}$$

Solutions: $$x = \frac{4 + 6}{2} = \frac{10}{2} = 5$$ $$x = \frac{4 - 6}{2} = \frac{-2}{2} = -1$$

The solutions are $x = 5$ and $x = -1$.

The Discriminant

The expression under the square root, $b^2 - 4ac$, has a special name: the discriminant. It's incredibly useful because it tells you how many solutions your equation has before you even finish solving it.

Here's how it works:

If $b^2 - 4ac > 0$ (positive): You have two different real solutions.

If $b^2 - 4ac = 0$: You have exactly one real solution (actually a repeated solution).

If $b^2 - 4ac < 0$ (negative): You have no real solutions. The solutions involve imaginary numbers, which you'll learn about later.

Example 4: Without solving, determine how many real solutions $x^2 + 6x + 9 = 0$ has.

Calculate the discriminant: $$b^2 - 4ac = 6^2 - 4(1)(9) = 36 - 36 = 0$$

Since the discriminant equals zero, there's exactly one real solution.

(If you solve it: $x = \frac{-6 \pm \sqrt{0}}{2} = \frac{-6}{2} = -3$. Just one solution.)

Example 5: How many real solutions does $x^2 + 2x + 5 = 0$ have?

Discriminant: $$b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$$

Since the discriminant is negative, there are no real solutions. You can't take the square root of a negative number (at least not with real numbers).

Example 6: How many real solutions does $2x^2 - 5x + 1 = 0$ have?

Discriminant: $$b^2 - 4ac = (-5)^2 - 4(2)(1) = 25 - 8 = 17$$

Since the discriminant is positive, there are two different real solutions.

When to Use the Quadratic Formula

You might be wondering when to use the quadratic formula versus other methods like factoring. Here's a practical guide:

Try factoring first if the equation looks simple and the numbers are small. Factoring is faster when it works. For example, $x^2 + 7x + 12 = 0$ factors nicely into $(x + 3)(x + 4) = 0$.

Use the quadratic formula when:

The equation doesn't factor easily
You have fractions or decimals as coefficients
You're not sure if it factors at all
You need decimal approximations rather than exact answers
You're in a hurry and don't want to waste time trying to factor

The beauty of the quadratic formula is that it always works. Always. Whether the equation factors or not, whether the coefficients are nice or messy, the formula will give you the answer.

Example 7: Solve $3x^2 + 2x - 4 = 0$.

This doesn't factor nicely, so use the formula: $$x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-4)}}{2(3)}$$ $$x = \frac{-2 \pm \sqrt{4 + 48}}{6}$$ $$x = \frac{-2 \pm \sqrt{52}}{6}$$

Simplify $\sqrt{52} = \sqrt{4 \cdot 13} = 2\sqrt{13}$: $$x = \frac{-2 \pm 2\sqrt{13}}{6}$$ $$x = \frac{-1 \pm \sqrt{13}}{3}$$

The exact solutions are $x = \frac{-1 + \sqrt{13}}{3}$ and $x = \frac{-1 - \sqrt{13}}{3}$.

If you need decimal approximations: $x \approx 0.87$ and $x \approx -1.54$.

Real-World Applications

Quadratic equations show up in physics, engineering, business, and many other fields. Here are some common situations:

Projectile Motion

When you throw a ball, its height over time follows a quadratic equation. The formula $h = -16t^2 + v_0t + h_0$ describes the height $h$ at time $t$, where $v_0$ is the initial velocity and $h_0$ is the initial height.

Example 8: A ball is thrown upward from a height of 5 feet with an initial velocity of 40 feet per second. When does it hit the ground?

The equation is $h = -16t^2 + 40t + 5$. The ball hits the ground when $h = 0$: $$-16t^2 + 40t + 5 = 0$$

Use the quadratic formula with $a = -16$, $b = 40$, $c = 5$: $$t = \frac{-40 \pm \sqrt{40^2 - 4(-16)(5)}}{2(-16)}$$ $$t = \frac{-40 \pm \sqrt{1600 + 320}}{-32}$$ $$t = \frac{-40 \pm \sqrt{1920}}{-32}$$ $$t = \frac{-40 \pm 43.82}{-32}$$

Two solutions: $$t = \frac{-40 + 43.82}{-32} \approx -0.12$$ (ignore, negative time doesn't make sense) $$t = \frac{-40 - 43.82}{-32} \approx 2.62$$

The ball hits the ground after about 2.62 seconds.

Area Problems

Many geometry problems lead to quadratic equations.

Example 9: A rectangle has a length that's 3 feet longer than its width. If the area is 40 square feet, what are the dimensions?

Let $w$ = width. Then length = $w + 3$.

Area equation: $$w(w + 3) = 40$$ $$w^2 + 3w = 40$$ $$w^2 + 3w - 40 = 0$$

This factors: $(w + 8)(w - 5) = 0$, so $w = -8$ or $w = 5$.

Since width can't be negative, $w = 5$ feet and length = $5 + 3 = 8$ feet.

Business Applications

Profit and revenue problems often involve quadratic equations.

Example 10: A company's profit in thousands of dollars is modeled by $P = -2x^2 + 20x - 32$, where $x$ is the number of units sold (in thousands). For what values of $x$ does the company break even (profit = 0)?

Set $P = 0$: $$-2x^2 + 20x - 32 = 0$$

Divide by -2 to simplify: $$x^2 - 10x + 16 = 0$$

This factors: $(x - 2)(x - 8) = 0$, so $x = 2$ or $x = 8$.

The company breaks even when selling 2,000 units or 8,000 units.

Practice Problems

Try these on your own, then check your answers.

Solve $x^2 + 4x - 12 = 0$
Solve $2x^2 + 5x - 3 = 0$
Solve $x^2 - 6x + 9 = 0$
Find the discriminant and determine the number of real solutions: $3x^2 - 2x + 1 = 0$
Solve $x^2 - 8x + 15 = 0$
Solve $4x^2 - 12x + 9 = 0$
A ball is thrown upward with an initial velocity of 32 feet per second from a height of 6 feet. Use $h = -16t^2 + 32t + 6$ to find when it hits the ground.
Solve $x^2 + x - 1 = 0$ (leave answer in exact form)

Check Your Work

$x = 2$ or $x = -6$
$x = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm 8}{2}$
$x = \frac{1}{2}$ or $x = -3$
$x = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm 7}{4}$
$x = 3$ (one solution)
$x = \frac{6 \pm \sqrt{36 - 36}}{2} = \frac{6}{2} = 3$
Discriminant = -8, zero real solutions
$b^2 - 4ac = 4 - 12 = -8$ (negative)
$x = 3$ or $x = 5$
Factors: $(x - 3)(x - 5) = 0$ or use formula
$x = \frac{3}{2}$ (one solution)
$x = \frac{12 \pm \sqrt{144 - 144}}{8} = \frac{12}{8} = \frac{3}{2}$
About 2.17 seconds
$t = \frac{-32 \pm \sqrt{1024 + 384}}{-32} = \frac{-32 \pm 37.53}{-32}$
Take positive solution: $t \approx 2.17$
$x = \frac{-1 \pm \sqrt{5}}{2}$
$x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$

Why This Matters

The quadratic formula is one of the most important tools in algebra. It gives you a systematic way to solve any quadratic equation, which means you can tackle problems in physics, engineering, business, and countless other fields.

More importantly, understanding the quadratic formula helps you see patterns in mathematics. The discriminant tells you something about the solutions before you even find them. The formula itself connects algebra, geometry (parabolas), and even calculus (where quadratics show up constantly).

Once you master this formula, you've unlocked a powerful problem-solving tool that will serve you well in higher math and beyond.