-2(x^2-3x-9)=3(x^2+2x+6)

Solution for -2(x^2-3x-9)=3(x^2+2x+6) equation:

-2(x^2-3x-9)=3(x^2+2x+6)

We move all terms to the left:

-2(x^2-3x-9)-(3(x^2+2x+6))=0

We multiply parentheses

-2x^2+6x-(3(x^2+2x+6))+18=0


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

3(x^2+2x+6)

We multiply parentheses

3x^2+6x+18

Back to the equation:

-(3x^2+6x+18)


We get rid of parentheses

-2x^2-3x^2+6x-6x-18+18=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:
$x=\frac{-b}{2a}=\frac{0}{-10}=0$