F(x)=2x^2+5x-20

Solution for F(x)=2x^2+5x-20 equation:

(F)=2F^2+5F-20

We move all terms to the left:

(F)-(2F^2+5F-20)=0

We get rid of parentheses

-2F^2+F-5F+20=0

We add all the numbers together, and all the variables

-2F^2-4F+20=0

a = -2; b = -4; c = +20;
Δ = b2-4ac
Δ = -42-4·(-2)·20
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{11}}{2*-2}=\frac{4-4\sqrt{11}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{11}}{2*-2}=\frac{4+4\sqrt{11}}{-4}
$