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

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

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

We move all terms to the left:

f2-(4f-2f-8)-(2f^2+4f-8)=0

We add all the numbers together, and all the variables

f2-(2f-8)-(2f^2+4f-8)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1f^2-6f+16=0

a = -1; b = -6; c = +16;
Δ = b2-4ac
Δ = -62-4·(-1)·16
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-10}{2*-1}=\frac{-4}{-2}
=+2
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+10}{2*-1}=\frac{16}{-2}
=-8
$