(158-4x)+(x^2-5x)=180

Solution for (158-4x)+(x^2-5x)=180 equation:

(158-4x)+(x^2-5x)=180

We move all terms to the left:

(158-4x)+(x^2-5x)-(180)=0

We add all the numbers together, and all the variables

(-4x+158)+(x^2-5x)-180=0

We get rid of parentheses

x^2-4x-5x+158-180=0

We add all the numbers together, and all the variables

x^2-9x-22=0

a = 1; b = -9; c = -22;
Δ = b2-4ac
Δ = -92-4·1·(-22)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-13}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+13}{2*1}=\frac{22}{2}
=11
$