r^2-12r-9=0

Solution for r^2-12r-9=0 equation:

r^2-12r-9=0

a = 1; b = -12; c = -9;
Δ = b2-4ac
Δ = -122-4·1·(-9)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{5}}{2*1}=\frac{12-6\sqrt{5}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{5}}{2*1}=\frac{12+6\sqrt{5}}{2}
$