3z(z+2)=180

Solution for 3z(z+2)=180 equation:

3z(z+2)=180

We move all terms to the left:

3z(z+2)-(180)=0

We multiply parentheses

3z^2+6z-180=0

a = 3; b = 6; c = -180;
Δ = b2-4ac
Δ = 62-4·3·(-180)
Δ = 2196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2196}=\sqrt{36*61}=\sqrt{36}*\sqrt{61}=6\sqrt{61}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{61}}{2*3}=\frac{-6-6\sqrt{61}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{61}}{2*3}=\frac{-6+6\sqrt{61}}{6}
$