z2+9z=18

Solution for z2+9z=18 equation:

z2+9z=18

We move all terms to the left:

z2+9z-(18)=0

We add all the numbers together, and all the variables

z^2+9z-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:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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