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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -((-4x+158)+(x^2-5x)), so:

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

We get rid of parentheses

x^2-4x-5x+158

We add all the numbers together, and all the variables

x^2-9x+158

Back to the equation:

-(x^2-9x+158)


We get rid of parentheses

-x^2+9x-158+180=0

We add all the numbers together, and all the variables

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