x^2-3x+x^2-x-20=180

Solution for x^2-3x+x^2-x-20=180 equation:

x^2-3x+x^2-x-20=180

We move all terms to the left:

x^2-3x+x^2-x-20-(180)=0

We add all the numbers together, and all the variables

2x^2-4x-200=0

a = 2; b = -4; c = -200;
Δ = b2-4ac
Δ = -42-4·2·(-200)
Δ = 1616
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:

$\sqrt{\Delta}=\sqrt{1616}=\sqrt{16*101}=\sqrt{16}*\sqrt{101}=4\sqrt{101}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{101}}{2*2}=\frac{4-4\sqrt{101}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{101}}{2*2}=\frac{4+4\sqrt{101}}{4}
$