7/10v+3/2=2+3/5v

Solution for 7/10v+3/2=2+3/5v equation:

7/10v+3/2=2+3/5v

We move all terms to the left:

7/10v+3/2-(2+3/5v)=0


Domain of the equation: 10v!=0

v!=0/10

v!=0

v∈R



Domain of the equation: 5v)!=0

v!=0/1

v!=0

v∈R


We add all the numbers together, and all the variables

7/10v-(3/5v+2)+3/2=0

We get rid of parentheses

7/10v-3/5v-2+3/2=0

We calculate fractions


750v^2/200v^2+140v/200v^2+(-120v)/200v^2-2=0

We multiply all the terms by the denominator


750v^2+140v+(-120v)-2*200v^2=0

Wy multiply elements

750v^2-400v^2+140v+(-120v)=0

We get rid of parentheses

750v^2-400v^2+140v-120v=0

We add all the numbers together, and all the variables

350v^2+20v=0

a = 350; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·350·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*350}=\frac{-40}{700}
=-2/35
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*350}=\frac{0}{700}
=0
$