18z^2+60z-50=0

Solution for 18z^2+60z-50=0 equation:

18z^2+60z-50=0

a = 18; b = 60; c = -50;
Δ = b2-4ac
Δ = 602-4·18·(-50)
Δ = 7200
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:

$\sqrt{\Delta}=\sqrt{7200}=\sqrt{3600*2}=\sqrt{3600}*\sqrt{2}=60\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-60\sqrt{2}}{2*18}=\frac{-60-60\sqrt{2}}{36}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+60\sqrt{2}}{2*18}=\frac{-60+60\sqrt{2}}{36}
$