1/3(12z)=6z

Solution for 1/3(12z)=6z equation:

1/3(12z)=6z

We move all terms to the left:

1/3(12z)-(6z)=0


Domain of the equation: 312z!=0

z!=0/312

z!=0

z∈R


We add all the numbers together, and all the variables

-6z+1/312z=0

We multiply all the terms by the denominator


-6z*312z+1=0

Wy multiply elements

-1872z^2+1=0

a = -1872; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-1872)·1
Δ = 7488
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:

$\sqrt{\Delta}=\sqrt{7488}=\sqrt{576*13}=\sqrt{576}*\sqrt{13}=24\sqrt{13}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{13}}{2*-1872}=\frac{0-24\sqrt{13}}{-3744}
=-\frac{24\sqrt{13}}{-3744}
=-\frac{\sqrt{13}}{-156}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{13}}{2*-1872}=\frac{0+24\sqrt{13}}{-3744}
=\frac{24\sqrt{13}}{-3744}
=\frac{\sqrt{13}}{-156}
$