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+6F+20=0

a = -2; b = 6; c = +20;
Δ = b2-4ac
Δ = 62-4·(-2)·20
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*-2}=\frac{-20}{-4}
=+5
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*-2}=\frac{8}{-4}
=-2
$