# Solve by completing square: 2x^2 - 6x - 8 = 0

#### marshall1432

##### Banned
Solve by completing the square:
2x'2'squared -6x-8=0

WORK DONE :

x2-3x-8=0

x2-3x-(3/2)2= -8=(3/2)2

(x-3/2)2=-8/2 +3/2
x-3/2 'plus or minus' -5/2

final answer that i got was ' x=i-5/2+3/2

#### Subhotosh Khan

##### Super Moderator
Staff member
Re: Solve by completing the square:

marshall1432 said:
Solve by completing the square:
2x'2'squared -6x-8=0

WORK DONE :

x2-3x-4=0

x2-3x-(3/2)2= -8=(3/2)2 <-- Why is 8 negative here

(x-3/2)2=-8/2 +3/2
x-3/2 'plus or minus' -5/2

final answer that i got was ' x=i-5/2+3/2

$$\displaystyle ax^2 \, + \, bx \, + c = 0$$

if

ac < 0 --> roots are real.
$$\displaystyle x^2\, - \, 3x \, = \, 4$$

$$\displaystyle (x\, - \frac{3}{2})^2 \, = \,4 + \frac{9}{4}$$

$$\displaystyle x\, =\, \frac{3}{2} \, \pm\sqrt{ \frac{25}{4}}\, = \, \frac{3\pm\ 5}{2} \, = \, ??$$

Staff member

#### marshall1432

##### Banned
I got the 3+-5
---------
2

BUT i DON'T KNOW WHERE TO GO FROM THERE. DO I NEED TO MULTIPLY EVERYTHING BY 2 AND DIVIDE OUT? THANKS

#### Subhotosh Khan

##### Super Moderator
Staff member
marshall1432 said:
I got the 3+-5
---------
2

No - you did not get it. You made multiple mistakes (like using '8' instead '4') and got to the number. But your process is so misguided - it is not even wrong!

Clean up your arithmatic first - looking at the procedure I posted.

Then think about the answer and how you can simplify from there.

BUT i DON'T KNOW WHERE TO GO FROM THERE.

DO I NEED TO MULTIPLY EVERYTHING BY 2 AND DIVIDE OUT?

I think there is a serious gap in your understanding of mathematical operations. My advice for you would be to get a tutor for face-to-face instructions.

THANKS