How do you solve the equation 2x^2-3x-3=0 by completing the square?

1 Answer
Nov 10, 2016

x=(3+-sqrt(15))/2

Explanation:

2x^2-3x-3=0

divide by 2 to make coefficient x^2=1

x^2-(3x)/2-3/2=0

(x^2-(3x)/2)-3/2=0

square half coefficient of x then add and subtract

(x^2-(3x)/2+(3/4)^2)-3/2-(3/4)^2=0

the first bracket is now a perfect square

(x-3/2)^2-3/2-9/4=0

(x-3/2)^2-15/4=0

now rearrange for x

(x-3/2)^2=15/4

x-3/2=+-sqrt(15/4)

x-3/2=+-sqrt(15)/2

x=3/2+-sqrt(15)/2

x=(3+-sqrt(15))/2