3z(z+11)=12

Solution for 3z(z+11)=12 equation:

3z(z+11)=12

We move all terms to the left:

3z(z+11)-(12)=0

We multiply parentheses

3z^2+33z-12=0

a = 3; b = 33; c = -12;
Δ = b2-4ac
Δ = 332-4·3·(-12)
Δ = 1233
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:

$\sqrt{\Delta}=\sqrt{1233}=\sqrt{9*137}=\sqrt{9}*\sqrt{137}=3\sqrt{137}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{137}}{2*3}=\frac{-33-3\sqrt{137}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{137}}{2*3}=\frac{-33+3\sqrt{137}}{6}
$