F(x)=4x^2+13x-33

Solution for F(x)=4x^2+13x-33 equation:

(F)=4F^2+13F-33

We move all terms to the left:

(F)-(4F^2+13F-33)=0

We get rid of parentheses

-4F^2+F-13F+33=0

We add all the numbers together, and all the variables

-4F^2-12F+33=0

a = -4; b = -12; c = +33;
Δ = b2-4ac
Δ = -122-4·(-4)·33
Δ = 672
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:

$\sqrt{\Delta}=\sqrt{672}=\sqrt{16*42}=\sqrt{16}*\sqrt{42}=4\sqrt{42}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{42}}{2*-4}=\frac{12-4\sqrt{42}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{42}}{2*-4}=\frac{12+4\sqrt{42}}{-8}
$