F(x)=3x^2+5x-36

Solution for F(x)=3x^2+5x-36 equation:

(F)=3F^2+5F-36

We move all terms to the left:

(F)-(3F^2+5F-36)=0

We get rid of parentheses

-3F^2+F-5F+36=0

We add all the numbers together, and all the variables

-3F^2-4F+36=0

a = -3; b = -4; c = +36;
Δ = b2-4ac
Δ = -42-4·(-3)·36
Δ = 448
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{7}}{2*-3}=\frac{4-8\sqrt{7}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{7}}{2*-3}=\frac{4+8\sqrt{7}}{-6}
$