(x^2-19x)+(80-2x)=180

Solution for (x^2-19x)+(80-2x)=180 equation:

(x^2-19x)+(80-2x)=180

We move all terms to the left:

(x^2-19x)+(80-2x)-(180)=0

We add all the numbers together, and all the variables

(x^2-19x)+(-2x+80)-180=0

We get rid of parentheses

x^2-19x-2x+80-180=0

We add all the numbers together, and all the variables

x^2-21x-100=0

a = 1; b = -21; c = -100;
Δ = b2-4ac
Δ = -212-4·1·(-100)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-29}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+29}{2*1}=\frac{50}{2}
=25
$