-1/3v-17/18=1/4v+1

Solution for -1/3v-17/18=1/4v+1 equation:

-1/3v-17/18=1/4v+1

We move all terms to the left:

-1/3v-17/18-(1/4v+1)=0


Domain of the equation: 3v!=0

v!=0/3

v!=0

v∈R



Domain of the equation: 4v+1)!=0

v∈R


We get rid of parentheses

-1/3v-1/4v-1-17/18=0

We calculate fractions


(-816v^2)/216v^2+(-72v)/216v^2+(-54v)/216v^2-1=0

We multiply all the terms by the denominator


(-816v^2)+(-72v)+(-54v)-1*216v^2=0

Wy multiply elements

(-816v^2)-216v^2+(-72v)+(-54v)=0

We get rid of parentheses

-816v^2-216v^2-72v-54v=0

We add all the numbers together, and all the variables

-1032v^2-126v=0

a = -1032; b = -126; c = 0;
Δ = b2-4ac
Δ = -1262-4·(-1032)·0
Δ = 15876
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15876}=126$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-126)-126}{2*-1032}=\frac{0}{-2064}
=0
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-126)+126}{2*-1032}=\frac{252}{-2064}
=-21/172
$