-3(z+4)+3/2z=z-7

Solution for -3(z+4)+3/2z=z-7 equation:

-3(z+4)+3/2z=z-7

We move all terms to the left:

-3(z+4)+3/2z-(z-7)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R


We multiply parentheses

-3z+3/2z-(z-7)-12=0

We get rid of parentheses

-3z+3/2z-z+7-12=0

We multiply all the terms by the denominator


-3z*2z-z*2z+7*2z-12*2z+3=0

Wy multiply elements

-6z^2-2z^2+14z-24z+3=0

We add all the numbers together, and all the variables

-8z^2-10z+3=0

a = -8; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·(-8)·3
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*-8}=\frac{-4}{-16}
=1/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*-8}=\frac{24}{-16}
=-1+1/2
$