k^2-12k=99

Solution for k^2-12k=99 equation:

k^2-12k=99

We move all terms to the left:

k^2-12k-(99)=0

a = 1; b = -12; c = -99;
Δ = b2-4ac
Δ = -122-4·1·(-99)
Δ = 540
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{15}}{2*1}=\frac{12-6\sqrt{15}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{15}}{2*1}=\frac{12+6\sqrt{15}}{2}
$