F(x)=10x2+5x-20

Solution for F(x)=10x2+5x-20 equation:

(F)=10F^2+5F-20

We move all terms to the left:

(F)-(10F^2+5F-20)=0

We get rid of parentheses

-10F^2+F-5F+20=0

We add all the numbers together, and all the variables

-10F^2-4F+20=0

a = -10; b = -4; c = +20;
Δ = b2-4ac
Δ = -42-4·(-10)·20
Δ = 816
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:

$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{51}}{2*-10}=\frac{4-4\sqrt{51}}{-20}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{51}}{2*-10}=\frac{4+4\sqrt{51}}{-20}
$