k^2+9k-9=0

Solution for k^2+9k-9=0 equation:

k^2+9k-9=0

a = 1; b = 9; c = -9;
Δ = b2-4ac
Δ = 92-4·1·(-9)
Δ = 117
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:

$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{13}}{2*1}=\frac{-9-3\sqrt{13}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{13}}{2*1}=\frac{-9+3\sqrt{13}}{2}
$