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

2 Answers
Aug 15, 2016

Answer:

#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#

Aug 15, 2016

Answer:

#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#