(r-3)(r+9)=11

Solution for (r-3)(r+9)=11 equation:

(r-3)(r+9)=11

We move all terms to the left:

(r-3)(r+9)-(11)=0

We multiply parentheses ..

(+r^2+9r-3r-27)-11=0

We get rid of parentheses

r^2+9r-3r-27-11=0

We add all the numbers together, and all the variables

r^2+6r-38=0

a = 1; b = 6; c = -38;
Δ = b2-4ac
Δ = 62-4·1·(-38)
Δ = 188
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:

$\sqrt{\Delta}=\sqrt{188}=\sqrt{4*47}=\sqrt{4}*\sqrt{47}=2\sqrt{47}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{47}}{2*1}=\frac{-6-2\sqrt{47}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{47}}{2*1}=\frac{-6+2\sqrt{47}}{2}
$