3/4z=1/2z+12

Solution for 3/4z=1/2z+12 equation:

3/4z=1/2z+12

We move all terms to the left:

3/4z-(1/2z+12)=0


Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R



Domain of the equation: 2z+12)!=0

z∈R


We get rid of parentheses

3/4z-1/2z-12=0

We calculate fractions


6z/8z^2+(-4z)/8z^2-12=0

We multiply all the terms by the denominator


6z+(-4z)-12*8z^2=0

Wy multiply elements

-96z^2+6z+(-4z)=0

We get rid of parentheses

-96z^2+6z-4z=0

We add all the numbers together, and all the variables

-96z^2+2z=0

a = -96; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-96)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-96}=\frac{-4}{-192}
=1/48
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-96}=\frac{0}{-192}
=0
$