2z^2+6z=180

Solution for 2z^2+6z=180 equation:

2z^2+6z=180

We move all terms to the left:

2z^2+6z-(180)=0

a = 2; b = 6; c = -180;
Δ = b2-4ac
Δ = 62-4·2·(-180)
Δ = 1476
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:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{41}}{2*2}=\frac{-6-6\sqrt{41}}{4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{41}}{2*2}=\frac{-6+6\sqrt{41}}{4}
$