3z(3z+17)=18

Solution for 3z(3z+17)=18 equation:

3z(3z+17)=18

We move all terms to the left:

3z(3z+17)-(18)=0

We multiply parentheses

9z^2+51z-18=0

a = 9; b = 51; c = -18;
Δ = b2-4ac
Δ = 512-4·9·(-18)
Δ = 3249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3249}=57$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-57}{2*9}=\frac{-108}{18}
=-6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+57}{2*9}=\frac{6}{18}
=1/3
$