z(3z+1)=14

Solution for z(3z+1)=14 equation:

z(3z+1)=14

We move all terms to the left:

z(3z+1)-(14)=0

We multiply parentheses

3z^2+z-14=0

a = 3; b = 1; c = -14;
Δ = b2-4ac
Δ = 12-4·3·(-14)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*3}=\frac{-14}{6}
=-2+1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*3}=\frac{12}{6}
=2
$