F(-5)=2x^2+4x-8

Solution for F(-5)=2x^2+4x-8 equation:

(-5)=2F^2+4F-8

We move all terms to the left:

(-5)-(2F^2+4F-8)=0

We add all the numbers together, and all the variables

-(2F^2+4F-8)-5=0

We get rid of parentheses

-2F^2-4F+8-5=0

We add all the numbers together, and all the variables

-2F^2-4F+3=0

a = -2; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·(-2)·3
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{10}}{2*-2}=\frac{4-2\sqrt{10}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{10}}{2*-2}=\frac{4+2\sqrt{10}}{-4}
$