4x-8=x^2-4

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Solution for 4x-8=x^2-4 equation: 



4x-8=x^2-4

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+4x-4=0

a = -1; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·(-1)·(-4)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

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

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