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

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

(2F)=F^2+3F

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1F^2-1F=0

a = -1; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-1)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-1}=\frac{0}{-2}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-1}=\frac{2}{-2}
=-1
$