F(x)=3x^2+24x+36

Solution for F(x)=3x^2+24x+36 equation:

(F)=3F^2+24F+36

We move all terms to the left:

(F)-(3F^2+24F+36)=0

We get rid of parentheses

-3F^2+F-24F-36=0

We add all the numbers together, and all the variables

-3F^2-23F-36=0

a = -3; b = -23; c = -36;
Δ = b2-4ac
Δ = -232-4·(-3)·(-36)
Δ = 97
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{-(-23)-\sqrt{97}}{2*-3}=\frac{23-\sqrt{97}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{97}}{2*-3}=\frac{23+\sqrt{97}}{-6}
$