F(0)=4x2-40x+5

Solution for F(0)=4x2-40x+5 equation:

(0)=4F^2-40F+5

We move all terms to the left:

(0)-(4F^2-40F+5)=0

We add all the numbers together, and all the variables

-(4F^2-40F+5)=0

We get rid of parentheses

-4F^2+40F-5=0

a = -4; b = 40; c = -5;
Δ = b2-4ac
Δ = 402-4·(-4)·(-5)
Δ = 1520
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:

$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{95}}{2*-4}=\frac{-40-4\sqrt{95}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{95}}{2*-4}=\frac{-40+4\sqrt{95}}{-8}
$