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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -4; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·(-4)·2
Δ = 57
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{57}}{2*-4}=\frac{-5-\sqrt{57}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{57}}{2*-4}=\frac{-5+\sqrt{57}}{-8}
$