Quadratic Formula help

[S_100]

M, g and h are constants
I don't understand how the equation
kx² -Mgx - Mgh = 0

Leads to the quadratic for x as seen in the solution.

Mg/2k (1 ± (1 + √ (4kh/Mg) )

----------------------------------------------------------

My attempt at a solution:
kx² -Mgx - Mgh = 0

Since M, g and h are constants
Quadratic formula can be applied to find x


x = (1/2a)* [-b ± √(b² - 4ac)]

kx² -Mgx - Mgh = 0



x² -(Mg/k)x - (Mgh/k) = 0

x = (1/2)* [(Mg/k)±√ ( (Mg/k)² - (4*-Mgh/k) )

x = (1/2)* [(Mg/k)±√ ( (Mg/k)² +(4Mgh/k) )



But from here I am lost on how to get to the last line of work shown in the solution:
Mg/2k (1 ± (1 + √ (4kh/Mg) )

 

[Deleted member 4993]

the quadratic solution for x as seen in the solution.

Mg/2k (1 ± (1 + √ (4kh/Mg) )

----------------------------------------------------------


kx² -Mgx - Mgh = 0

Here we have:

a = k

b = -Mg

c = -Mgh

Then

\(\displaystyle x = \frac{Mg}{2k} \pm \dfrac{\sqrt{M^2g^2 + 4*k*Mgh}}{2*k}\)

\(\displaystyle x = \frac{Mg}{2k} \pm \dfrac{M*g*\sqrt{1 + \frac{4*kh}{Mg}}}{2*k}\)

Continue.......

 

[Steven G]

If you are going to use the quadratic formula PLEASE do not start off by dividing by the coefficient of x^2 (k in this case).

 

[Otis]

If you are going to use the quadratic formula PLEASE do not start off by dividing by the coefficient of x^2 (k in this case).

Agree.

@S_100 \(\;\) Dividing by k (the leading coefficient) would be the first step, if we were solving by completing the square instead of using the quadratic formula.