F(x)=2x^2-3x-8

Solution for F(x)=2x^2-3x-8 equation:

(F)=2F^2-3F-8

We move all terms to the left:

(F)-(2F^2-3F-8)=0

We get rid of parentheses

-2F^2+F+3F+8=0

We add all the numbers together, and all the variables

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

a = -2; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·(-2)·8
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{5}}{2*-2}=\frac{-4-4\sqrt{5}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{5}}{2*-2}=\frac{-4+4\sqrt{5}}{-4}
$