F(x)=5x^2+40x+9

Solution for F(x)=5x^2+40x+9 equation:

(F)=5F^2+40F+9

We move all terms to the left:

(F)-(5F^2+40F+9)=0

We get rid of parentheses

-5F^2+F-40F-9=0

We add all the numbers together, and all the variables

-5F^2-39F-9=0

a = -5; b = -39; c = -9;
Δ = b2-4ac
Δ = -392-4·(-5)·(-9)
Δ = 1341
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:

$\sqrt{\Delta}=\sqrt{1341}=\sqrt{9*149}=\sqrt{9}*\sqrt{149}=3\sqrt{149}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-3\sqrt{149}}{2*-5}=\frac{39-3\sqrt{149}}{-10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+3\sqrt{149}}{2*-5}=\frac{39+3\sqrt{149}}{-10}
$