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

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

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

We move all terms to the left:

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


Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R



Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R


We add all the numbers together, and all the variables

3/4z-1/2z-16=0

We calculate fractions


6z/8z^2+(-4z)/8z^2-16=0

We multiply all the terms by the denominator


6z+(-4z)-16*8z^2=0

Wy multiply elements

-128z^2+6z+(-4z)=0

We get rid of parentheses

-128z^2+6z-4z=0

We add all the numbers together, and all the variables

-128z^2+2z=0

a = -128; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-128)·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*-128}=\frac{-4}{-256}
=1/64
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-128}=\frac{0}{-256}
=0
$