F(0)=5x^2+4x

Solution for F(0)=5x^2+4x equation:

(0)=5F^2+4F

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-5F^2-4F=0

a = -5; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-5)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-5}=\frac{0}{-10}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-5}=\frac{8}{-10}
=-4/5
$