f^2-(4f-2f^2-8)=2f2+4f-8

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Solution for f^2-(4f-2f^2-8)=2f2+4f-8 equation: 



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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

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