3(2z+2)-4z(z-15)=18

Solution for 3(2z+2)-4z(z-15)=18 equation:

3(2z+2)-4z(z-15)=18

We move all terms to the left:

3(2z+2)-4z(z-15)-(18)=0

We multiply parentheses

-4z^2+6z+60z+6-18=0

We add all the numbers together, and all the variables

-4z^2+66z-12=0

a = -4; b = 66; c = -12;
Δ = b2-4ac
Δ = 662-4·(-4)·(-12)
Δ = 4164
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:

$\sqrt{\Delta}=\sqrt{4164}=\sqrt{4*1041}=\sqrt{4}*\sqrt{1041}=2\sqrt{1041}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(66)-2\sqrt{1041}}{2*-4}=\frac{-66-2\sqrt{1041}}{-8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(66)+2\sqrt{1041}}{2*-4}=\frac{-66+2\sqrt{1041}}{-8}
$