F(2c)=5(2c)^2+4(2c)

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

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

We move all terms to the left:

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

We get rid of parentheses

-52F^2+2F-42F=0

We add all the numbers together, and all the variables

-52F^2-40F=0

a = -52; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·(-52)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*-52}=\frac{0}{-104}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*-52}=\frac{80}{-104}
=-10/13
$