F(x)=2x^2+10x-1.02

Solution for F(x)=2x^2+10x-1.02 equation:

(F)=2F^2+10F-1.02

We move all terms to the left:

(F)-(2F^2+10F-1.02)=0

We get rid of parentheses

-2F^2+F-10F+1.02=0

We add all the numbers together, and all the variables

-2F^2-9F+1.02=0

a = -2; b = -9; c = +1.02;
Δ = b2-4ac
Δ = -92-4·(-2)·1.02
Δ = 89.16
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{-(-9)-\sqrt{89.16}}{2*-2}=\frac{9-\sqrt{89.16}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{89.16}}{2*-2}=\frac{9+\sqrt{89.16}}{-4}
$