F(x)=x^2+4x-45

Solution for F(x)=x^2+4x-45 equation:

(F)=F^2+4F-45

We move all terms to the left:

(F)-(F^2+4F-45)=0

We get rid of parentheses

-F^2+F-4F+45=0

We add all the numbers together, and all the variables

-1F^2-3F+45=0

a = -1; b = -3; c = +45;
Δ = b2-4ac
Δ = -32-4·(-1)·45
Δ = 189
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:

$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{21}}{2*-1}=\frac{3-3\sqrt{21}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{21}}{2*-1}=\frac{3+3\sqrt{21}}{-2}
$