2x^2+x(8-x)=-16

Solution for 2x^2+x(8-x)=-16 equation:

2x^2+x(8-x)=-16

We move all terms to the left:

2x^2+x(8-x)-(-16)=0

We add all the numbers together, and all the variables

2x^2+x(-1x+8)-(-16)=0

We add all the numbers together, and all the variables

2x^2+x(-1x+8)+16=0

We multiply parentheses

2x^2-1x^2+8x+16=0

We add all the numbers together, and all the variables

x^2+8x+16=0

a = 1; b = 8; c = +16;
Δ = b2-4ac
Δ = 82-4·1·16
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-8}{2}=-4$