F(x)=x2+11x+20

Solution for F(x)=x2+11x+20 equation:

(F)=F2+11F+20

We move all terms to the left:

(F)-(F2+11F+20)=0

We add all the numbers together, and all the variables

-(+F^2+11F+20)+F=0

We get rid of parentheses

-F^2-11F+F-20=0

We add all the numbers together, and all the variables

-1F^2-10F-20=0

a = -1; b = -10; c = -20;
Δ = b2-4ac
Δ = -102-4·(-1)·(-20)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{5}}{2*-1}=\frac{10-2\sqrt{5}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{5}}{2*-1}=\frac{10+2\sqrt{5}}{-2}
$