2(3-z)=16-2(3/2z)

Solution for 2(3-z)=16-2(3/2z) equation:

2(3-z)=16-2(3/2z)

We move all terms to the left:

2(3-z)-(16-2(3/2z))=0


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

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

2(-1z+3)-(16-2(+3/2z))=0

We multiply parentheses

-2z-(16-2(+3/2z))+6=0

We multiply all the terms by the denominator


-2z*2z))-(16-2(+6*2z))+3=0

Wy multiply elements

-4z^2+12z=0

a = -4; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-4)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-4}=\frac{-24}{-8}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-4}=\frac{0}{-8}
=0
$