(r-9)r=11

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

(r-9)r=11

We move all terms to the left:

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

We multiply parentheses

r^2-9r-11=0

a = 1; b = -9; c = -11;
Δ = b2-4ac
Δ = -92-4·1·(-11)
Δ = 125
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:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-5\sqrt{5}}{2*1}=\frac{9-5\sqrt{5}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+5\sqrt{5}}{2*1}=\frac{9+5\sqrt{5}}{2}
$