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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -3; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·(-3)·1
Δ = 21
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{-(-3)-\sqrt{21}}{2*-3}=\frac{3-\sqrt{21}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{21}}{2*-3}=\frac{3+\sqrt{21}}{-6}
$