(158-4x)+(x^2-32)=180

Solution for (158-4x)+(x^2-32)=180 equation:

(158-4x)+(x^2-32)=180

We move all terms to the left:

(158-4x)+(x^2-32)-(180)=0

We add all the numbers together, and all the variables

(-4x+158)+(x^2-32)-180=0

We get rid of parentheses

x^2-4x+158-32-180=0

We add all the numbers together, and all the variables

x^2-4x-54=0

a = 1; b = -4; c = -54;
Δ = b2-4ac
Δ = -42-4·1·(-54)
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{58}}{2*1}=\frac{4-2\sqrt{58}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{58}}{2*1}=\frac{4+2\sqrt{58}}{2}
$