F(x)=x^2+16x-40

Solution for F(x)=x^2+16x-40 equation:

(F)=F^2+16F-40

We move all terms to the left:

(F)-(F^2+16F-40)=0

We get rid of parentheses

-F^2+F-16F+40=0

We add all the numbers together, and all the variables

-1F^2-15F+40=0

a = -1; b = -15; c = +40;
Δ = b2-4ac
Δ = -152-4·(-1)·40
Δ = 385
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{-(-15)-\sqrt{385}}{2*-1}=\frac{15-\sqrt{385}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{385}}{2*-1}=\frac{15+\sqrt{385}}{-2}
$