r^2-20r=31

Solution for r^2-20r=31 equation:

r^2-20r=31

We move all terms to the left:

r^2-20r-(31)=0

a = 1; b = -20; c = -31;
Δ = b2-4ac
Δ = -202-4·1·(-31)
Δ = 524
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{524}=\sqrt{4*131}=\sqrt{4}*\sqrt{131}=2\sqrt{131}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{131}}{2*1}=\frac{20-2\sqrt{131}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{131}}{2*1}=\frac{20+2\sqrt{131}}{2}
$