(r-5)(r-9)=12

Solution for (r-5)(r-9)=12 equation:

(r-5)(r-9)=12

We move all terms to the left:

(r-5)(r-9)-(12)=0

We multiply parentheses ..

(+r^2-9r-5r+45)-12=0

We get rid of parentheses

r^2-9r-5r+45-12=0

We add all the numbers together, and all the variables

r^2-14r+33=0

a = 1; b = -14; c = +33;
Δ = b2-4ac
Δ = -142-4·1·33
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-8}{2*1}=\frac{6}{2}
=3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+8}{2*1}=\frac{22}{2}
=11
$