(r-2)(r+20)=4

Solution for (r-2)(r+20)=4 equation:

(r-2)(r+20)=4

We move all terms to the left:

(r-2)(r+20)-(4)=0

We multiply parentheses ..

(+r^2+20r-2r-40)-4=0

We get rid of parentheses

r^2+20r-2r-40-4=0

We add all the numbers together, and all the variables

r^2+18r-44=0

a = 1; b = 18; c = -44;
Δ = b2-4ac
Δ = 182-4·1·(-44)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-10\sqrt{5}}{2*1}=\frac{-18-10\sqrt{5}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+10\sqrt{5}}{2*1}=\frac{-18+10\sqrt{5}}{2}
$