2z^2+9z-18=0

Solution for 2z^2+9z-18=0 equation:

2z^2+9z-18=0

a = 2; b = 9; c = -18;
Δ = b2-4ac
Δ = 92-4·2·(-18)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*2}=\frac{-24}{4}
=-6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*2}=\frac{6}{4}
=1+1/2
$