20+4x=x^2+8

Solution for 20+4x=x^2+8 equation:

20+4x=x^2+8

We move all terms to the left:

20+4x-(x^2+8)=0

We get rid of parentheses

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