(x+4)2+20=x2-18

Solution for (x+4)2+20=x2-18 equation:

(x+4)2+20=x2-18

We move all terms to the left:

(x+4)2+20-(x2-18)=0

We add all the numbers together, and all the variables

-(+x^2-18)+(x+4)2+20=0

We multiply parentheses

-(+x^2-18)+2x+8+20=0

We get rid of parentheses

-x^2+2x+18+8+20=0

We add all the numbers together, and all the variables

-1x^2+2x+46=0

a = -1; b = 2; c = +46;
Δ = b2-4ac
Δ = 22-4·(-1)·46
Δ = 188
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:

$\sqrt{\Delta}=\sqrt{188}=\sqrt{4*47}=\sqrt{4}*\sqrt{47}=2\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{47}}{2*-1}=\frac{-2-2\sqrt{47}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{47}}{2*-1}=\frac{-2+2\sqrt{47}}{-2}
$