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

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

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

We move all terms to the left:

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


Domain of the equation: 2v!=0

v!=0/2

v!=0

v∈R



Domain of the equation: 2v)!=0

v!=0/1

v!=0

v∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

3/2v-(+4v-5/2v)+7=0

We get rid of parentheses

3/2v-4v+5/2v+7=0

We multiply all the terms by the denominator


-4v*2v+7*2v+3+5=0

We add all the numbers together, and all the variables

-4v*2v+7*2v+8=0

Wy multiply elements

-8v^2+14v+8=0

a = -8; b = 14; c = +8;
Δ = b2-4ac
Δ = 142-4·(-8)·8
Δ = 452
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{452}=\sqrt{4*113}=\sqrt{4}*\sqrt{113}=2\sqrt{113}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{113}}{2*-8}=\frac{-14-2\sqrt{113}}{-16}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{113}}{2*-8}=\frac{-14+2\sqrt{113}}{-16}
$