F(x)=5x^2+40x-3

Solution for F(x)=5x^2+40x-3 equation:

(F)=5F^2+40F-3

We move all terms to the left:

(F)-(5F^2+40F-3)=0

We get rid of parentheses

-5F^2+F-40F+3=0

We add all the numbers together, and all the variables

-5F^2-39F+3=0

a = -5; b = -39; c = +3;
Δ = b2-4ac
Δ = -392-4·(-5)·3
Δ = 1581
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{-(-39)-\sqrt{1581}}{2*-5}=\frac{39-\sqrt{1581}}{-10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1581}}{2*-5}=\frac{39+\sqrt{1581}}{-10}
$