7/10v+32=2+3/5v

Solution for 7/10v+32=2+3/5v equation:

7/10v+32=2+3/5v

We move all terms to the left:

7/10v+32-(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)+32=0

We get rid of parentheses

7/10v-3/5v-2+32=0

We calculate fractions


35v/50v^2+(-30v)/50v^2-2+32=0

We add all the numbers together, and all the variables

35v/50v^2+(-30v)/50v^2+30=0

We multiply all the terms by the denominator


35v+(-30v)+30*50v^2=0

Wy multiply elements

1500v^2+35v+(-30v)=0

We get rid of parentheses

1500v^2+35v-30v=0

We add all the numbers together, and all the variables

1500v^2+5v=0

a = 1500; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·1500·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*1500}=\frac{-10}{3000}
=-1/300
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*1500}=\frac{0}{3000}
=0
$