F(5x)=3x^2+1

Solution for F(5x)=3x^2+1 equation:

(5F)=3F^2+1

We move all terms to the left:

(5F)-(3F^2+1)=0

We get rid of parentheses

-3F^2+5F-1=0

a = -3; b = 5; c = -1;
Δ = b2-4ac
Δ = 52-4·(-3)·(-1)
Δ = 13
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{13}}{2*-3}=\frac{-5-\sqrt{13}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{13}}{2*-3}=\frac{-5+\sqrt{13}}{-6}
$