F(x^2+3x)=2x^2+6x-3

Solution for F(x^2+3x)=2x^2+6x-3 equation:

(F^2+3F)=2F^2+6F-3

We move all terms to the left:

(F^2+3F)-(2F^2+6F-3)=0

We get rid of parentheses

F^2-2F^2+3F-6F+3=0

We add all the numbers together, and all the variables

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

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