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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2F^2-9F+5=0

a = -2; b = -9; c = +5;
Δ = b2-4ac
Δ = -92-4·(-2)·5
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-11}{2*-2}=\frac{-2}{-4}
=1/2
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+11}{2*-2}=\frac{20}{-4}
=-5
$