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

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

a = 1; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·1·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*1}=\frac{0}{2}
=0
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*1}=\frac{16}{2}
=8
$