z(3z+1)=140

Solution for z(3z+1)=140 equation:

z(3z+1)=140

We move all terms to the left:

z(3z+1)-(140)=0

We multiply parentheses

3z^2+z-140=0

a = 3; b = 1; c = -140;
Δ = b2-4ac
Δ = 12-4·3·(-140)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-41}{2*3}=\frac{-42}{6}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+41}{2*3}=\frac{40}{6}
=6+2/3
$