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

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

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

We move all terms to the left:

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


Domain of the equation: 2)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

3z^2-2z+2-4=0

We add all the numbers together, and all the variables

3z^2-2z-2=0

a = 3; b = -2; c = -2;
Δ = b2-4ac
Δ = -22-4·3·(-2)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*3}=\frac{2-2\sqrt{7}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*3}=\frac{2+2\sqrt{7}}{6}
$