3(x^2-2x-4)=-2(x^2+3x+6)

Solution for 3(x^2-2x-4)=-2(x^2+3x+6) equation:

3(x^2-2x-4)=-2(x^2+3x+6)

We move all terms to the left:

3(x^2-2x-4)-(-2(x^2+3x+6))=0

We multiply parentheses

3x^2-6x-(-2(x^2+3x+6))-12=0


We calculate terms in parentheses: -(-2(x^2+3x+6)), so:

-2(x^2+3x+6)

We multiply parentheses

-2x^2-6x-12

Back to the equation:

-(-2x^2-6x-12)


We get rid of parentheses

3x^2+2x^2+6x-6x+12-12=0

We add all the numbers together, and all the variables

5x^2=0

a = 5; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·5·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{10}=0$