3z2+6z=24

Solution for 3z2+6z=24 equation:

3z^2+6z=24

We move all terms to the left:

3z^2+6z-(24)=0

a = 3; b = 6; c = -24;
Δ = b2-4ac
Δ = 62-4·3·(-24)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*3}=\frac{-24}{6}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*3}=\frac{12}{6}
=2
$