Question

How do you solve `2x^2-2x-12=0`?

Answer 1

`x= 3 or x = -2`

Explanation:

All the numbers in the equation are even, so divide through by 2.

`x^2 - x - 6 = 0" find factors"`

The factors of 6 which subtract to make 1 are 2 and 2.

`(x-3)(x+2) = 0`

If the answer to multiplication is 0, then one of the factors must be 0,

If `x-3 = 0, "then " x=3`

If `x+2 = 0, "then " x = -2`

Answer 2

`x=-2,x=3`

Explanation:

The first step is to take out a `color(blue)"common factor"` of 2.

`rArr2(x^2-x-6)=0`

We now require to factorise the `color(blue)"quadratic"` in the
bracket.

Find the factors of - 6 which sum to - 1.

These are - 3 and +2

`rArr2x^2-2x-12=2(x-3)(x+2)=0`

Either of the factors (x - 3) , (x +2) can equal zero.

solve : `x-3=0rArrx=3`

solve : `x+2=0rArrx=-2`