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

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

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

We move all terms to the left:

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

We get rid of parentheses

-55F^2+5F-25F+4=0

We add all the numbers together, and all the variables

-55F^2-20F+4=0

a = -55; b = -20; c = +4;
Δ = b2-4ac
Δ = -202-4·(-55)·4
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16\sqrt{5}}{2*-55}=\frac{20-16\sqrt{5}}{-110}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16\sqrt{5}}{2*-55}=\frac{20+16\sqrt{5}}{-110}
$