x^2-x-20=03x^2-21x=0

Solution for x^2-x-20=03x^2-21x=0 equation:

x^2-x-20=03x^2-21x=0

We move all terms to the left:

x^2-x-20-(03x^2-21x)=0

We add all the numbers together, and all the variables

x^2-1x-(03x^2-21x)-20=0

We get rid of parentheses

x^2-03x^2-1x+21x-20=0

We add all the numbers together, and all the variables

-2x^2+20x-20=0

a = -2; b = 20; c = -20;
Δ = b2-4ac
Δ = 202-4·(-2)·(-20)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{15}}{2*-2}=\frac{-20-4\sqrt{15}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{15}}{2*-2}=\frac{-20+4\sqrt{15}}{-4}
$