6(z-2)=4(z-3)z

Solution for 6(z-2)=4(z-3)z equation:

6(z-2)=4(z-3)z

We move all terms to the left:

6(z-2)-(4(z-3)z)=0

We multiply parentheses

6z-(4(z-3)z)-12=0


We calculate terms in parentheses: -(4(z-3)z), so:

4(z-3)z

We multiply parentheses

4z^2-12z

Back to the equation:

-(4z^2-12z)


We get rid of parentheses

-4z^2+6z+12z-12=0

We add all the numbers together, and all the variables

-4z^2+18z-12=0

a = -4; b = 18; c = -12;
Δ = b2-4ac
Δ = 182-4·(-4)·(-12)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{33}}{2*-4}=\frac{-18-2\sqrt{33}}{-8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{33}}{2*-4}=\frac{-18+2\sqrt{33}}{-8}
$