F(5x)=x^2-4

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

(5F)=F^2-4

We move all terms to the left:

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

We get rid of parentheses

-F^2+5F+4=0

We add all the numbers together, and all the variables

-1F^2+5F+4=0

a = -1; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·(-1)·4
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{41}}{2*-1}=\frac{-5-\sqrt{41}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{41}}{2*-1}=\frac{-5+\sqrt{41}}{-2}
$