3z2+6z-5=0

Solution for 3z2+6z-5=0 equation:

3z^2+6z-5=0

a = 3; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·3·(-5)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{6}}{2*3}=\frac{-6-4\sqrt{6}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{6}}{2*3}=\frac{-6+4\sqrt{6}}{6}
$