3z^2+60z=39

Solution for 3z^2+60z=39 equation:

3z^2+60z=39

We move all terms to the left:

3z^2+60z-(39)=0

a = 3; b = 60; c = -39;
Δ = b2-4ac
Δ = 602-4·3·(-39)
Δ = 4068
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:

$\sqrt{\Delta}=\sqrt{4068}=\sqrt{36*113}=\sqrt{36}*\sqrt{113}=6\sqrt{113}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-6\sqrt{113}}{2*3}=\frac{-60-6\sqrt{113}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+6\sqrt{113}}{2*3}=\frac{-60+6\sqrt{113}}{6}
$