(r-3)(r+9)=44

Solution for (r-3)(r+9)=44 equation:

(r-3)(r+9)=44

We move all terms to the left:

(r-3)(r+9)-(44)=0

We multiply parentheses ..

(+r^2+9r-3r-27)-44=0

We get rid of parentheses

r^2+9r-3r-27-44=0

We add all the numbers together, and all the variables

r^2+6r-71=0

a = 1; b = 6; c = -71;
Δ = b2-4ac
Δ = 62-4·1·(-71)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-8\sqrt{5}}{2*1}=\frac{-6-8\sqrt{5}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+8\sqrt{5}}{2*1}=\frac{-6+8\sqrt{5}}{2}
$