-3/2v+3/2=-4/5v-3

Solution for -3/2v+3/2=-4/5v-3 equation:

-3/2v+3/2=-4/5v-3

We move all terms to the left:

-3/2v+3/2-(-4/5v-3)=0


Domain of the equation: 2v!=0

v!=0/2

v!=0

v∈R



Domain of the equation: 5v-3)!=0

v∈R


We get rid of parentheses

-3/2v+4/5v+3+3/2=0

We calculate fractions


(-15v)/40v^2+32v/40v^2+15v/40v^2+3=0

We multiply all the terms by the denominator


(-15v)+32v+15v+3*40v^2=0

We add all the numbers together, and all the variables

47v+(-15v)+3*40v^2=0

Wy multiply elements

120v^2+47v+(-15v)=0

We get rid of parentheses

120v^2+47v-15v=0

We add all the numbers together, and all the variables

120v^2+32v=0

a = 120; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·120·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*120}=\frac{-64}{240}
=-4/15
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*120}=\frac{0}{240}
=0
$