4z^2+3z-6=12

Solution for 4z^2+3z-6=12 equation:

4z^2+3z-6=12

We move all terms to the left:

4z^2+3z-6-(12)=0

We add all the numbers together, and all the variables

4z^2+3z-18=0

a = 4; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·4·(-18)
Δ = 297
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:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{33}}{2*4}=\frac{-3-3\sqrt{33}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{33}}{2*4}=\frac{-3+3\sqrt{33}}{8}
$