-4z(z-12)=42

Solution for -4z(z-12)=42 equation:

-4z(z-12)=42

We move all terms to the left:

-4z(z-12)-(42)=0

We multiply parentheses

-4z^2+48z-42=0

a = -4; b = 48; c = -42;
Δ = b2-4ac
Δ = 482-4·(-4)·(-42)
Δ = 1632
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:

$\sqrt{\Delta}=\sqrt{1632}=\sqrt{16*102}=\sqrt{16}*\sqrt{102}=4\sqrt{102}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-4\sqrt{102}}{2*-4}=\frac{-48-4\sqrt{102}}{-8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+4\sqrt{102}}{2*-4}=\frac{-48+4\sqrt{102}}{-8}
$