k^2+9k=18

Solution for k^2+9k=18 equation:

k^2+9k=18

We move all terms to the left:

k^2+9k-(18)=0

a = 1; b = 9; c = -18;
Δ = b2-4ac
Δ = 92-4·1·(-18)
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{17}}{2*1}=\frac{-9-3\sqrt{17}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{17}}{2*1}=\frac{-9+3\sqrt{17}}{2}
$