F=2x^2+5x-10

Solution for F=2x^2+5x-10 equation:

=2F^2+5F-10

We move all terms to the left:

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

We get rid of parentheses

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

a = -2; b = -5; c = +10;
Δ = b2-4ac
Δ = -52-4·(-2)·10
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{105}}{2*-2}=\frac{5-\sqrt{105}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{105}}{2*-2}=\frac{5+\sqrt{105}}{-4}
$