2x+20=(x^2)+(6x+8)

Solution for 2x+20=(x^2)+(6x+8) equation:

2x+20=(x^2)+(6x+8)

We move all terms to the left:

2x+20-((x^2)+(6x+8))=0


We calculate terms in parentheses: -(x^2+(6x+8)), so:

x^2+(6x+8)

We get rid of parentheses

x^2+6x+8

Back to the equation:

-(x^2+6x+8)


We get rid of parentheses

-x^2+2x-6x-8+20=0

We add all the numbers together, and all the variables

-1x^2-4x+12=0

a = -1; b = -4; c = +12;
Δ = b2-4ac
Δ = -42-4·(-1)·12
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-1}=\frac{12}{-2}
=-6
$