F(x)=10x^2+x-2

Solution for F(x)=10x^2+x-2 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-10F^2+2=0

a = -10; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-10)·2
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{-(0)-4\sqrt{5}}{2*-10}=\frac{0-4\sqrt{5}}{-20}
=-\frac{4\sqrt{5}}{-20}
=-\frac{\sqrt{5}}{-5}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-10}=\frac{0+4\sqrt{5}}{-20}
=\frac{4\sqrt{5}}{-20}
=\frac{\sqrt{5}}{-5}
$