2z(3z+17)=12

Solution for 2z(3z+17)=12 equation:

2z(3z+17)=12

We move all terms to the left:

2z(3z+17)-(12)=0

We multiply parentheses

6z^2+34z-12=0

a = 6; b = 34; c = -12;
Δ = b2-4ac
Δ = 342-4·6·(-12)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-38}{2*6}=\frac{-72}{12}
=-6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+38}{2*6}=\frac{4}{12}
=1/3
$