2+4x=x^2-8

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

2+4x=x^2-8

We move all terms to the left:

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

We get rid of parentheses

-x^2+4x+8+2=0

We add all the numbers together, and all the variables

-1x^2+4x+10=0

a = -1; b = 4; c = +10;
Δ = b2-4ac
Δ = 42-4·(-1)·10
Δ = 56
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{14}}{2*-1}=\frac{-4-2\sqrt{14}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{14}}{2*-1}=\frac{-4+2\sqrt{14}}{-2}
$