(47-2x)+(x^2+32)=180

Solution for (47-2x)+(x^2+32)=180 equation:

(47-2x)+(x^2+32)=180

We move all terms to the left:

(47-2x)+(x^2+32)-(180)=0

We add all the numbers together, and all the variables

(-2x+47)+(x^2+32)-180=0

We get rid of parentheses

x^2-2x+47+32-180=0

We add all the numbers together, and all the variables

x^2-2x-101=0

a = 1; b = -2; c = -101;
Δ = b2-4ac
Δ = -22-4·1·(-101)
Δ = 408
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:

$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{102}}{2*1}=\frac{2-2\sqrt{102}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{102}}{2*1}=\frac{2+2\sqrt{102}}{2}
$