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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We calculate fractions


500v^2/450v^2+315v/450v^2+(-270v)/450v^2-2=0

We multiply all the terms by the denominator


500v^2+315v+(-270v)-2*450v^2=0

Wy multiply elements

500v^2-900v^2+315v+(-270v)=0

We get rid of parentheses

500v^2-900v^2+315v-270v=0

We add all the numbers together, and all the variables

-400v^2+45v=0

a = -400; b = 45; c = 0;
Δ = b2-4ac
Δ = 452-4·(-400)·0
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-45}{2*-400}=\frac{-90}{-800}
=9/80
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+45}{2*-400}=\frac{0}{-800}
=0
$