20-2x-x^2=x+2

Solution for 20-2x-x^2=x+2 equation:

20-2x-x^2=x+2

We move all terms to the left:

20-2x-x^2-(x+2)=0

We add all the numbers together, and all the variables

-1x^2-2x-(x+2)+20=0

We get rid of parentheses

-1x^2-2x-x-2+20=0

We add all the numbers together, and all the variables

-1x^2-3x+18=0

a = -1; b = -3; c = +18;
Δ = b2-4ac
Δ = -32-4·(-1)·18
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*-1}=\frac{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*-1}=\frac{12}{-2}
=-6
$