x^2-(4x-2x^2-8)=2x^2+4x-8

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Solution for x^2-(4x-2x^2-8)=2x^2+4x-8 equation: 



x^2-(4x-2x^2-8)=2x^2+4x-8

We move all terms to the left:

x^2-(4x-2x^2-8)-(2x^2+4x-8)=0

We get rid of parentheses

x^2+2x^2-2x^2-4x-4x+8+8=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:
$x=\frac{-b}{2a}=\frac{8}{2}=4$

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