F(x)=x^2+22x+40

Solution for F(x)=x^2+22x+40 equation:

(F)=F^2+22F+40

We move all terms to the left:

(F)-(F^2+22F+40)=0

We get rid of parentheses

-F^2+F-22F-40=0

We add all the numbers together, and all the variables

-1F^2-21F-40=0

a = -1; b = -21; c = -40;
Δ = b2-4ac
Δ = -212-4·(-1)·(-40)
Δ = 281
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{-(-21)-\sqrt{281}}{2*-1}=\frac{21-\sqrt{281}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{281}}{2*-1}=\frac{21+\sqrt{281}}{-2}
$