-0.16+0.2v=1/9v-0.5v

Solution for -0.16+0.2v=1/9v-0.5v equation:

-0.16+0.2v=1/9v-0.5v

We move all terms to the left:

-0.16+0.2v-(1/9v-0.5v)=0


Domain of the equation: 9v-0.5v)!=0

v∈R


We add all the numbers together, and all the variables

0.2v-(-0.5v+1/9v)-0.16=0

We get rid of parentheses

0.2v+0.5v-1/9v-0.16=0

We multiply all the terms by the denominator


(0.2v)*9v+(0.5v)*9v-(0.16)*9v-1=0

We add all the numbers together, and all the variables

(+0.2v)*9v+(+0.5v)*9v-(0.16)*9v-1=0

We multiply parentheses

0v^2+0v^2-1.44v-1=0

We add all the numbers together, and all the variables

2v^2-1.44v-1=0

a = 2; b = -1.44; c = -1;
Δ = b2-4ac
Δ = -1.442-4·2·(-1)
Δ = 10.0736
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1.44)-\sqrt{10.0736}}{2*2}=\frac{1.44-\sqrt{10.0736}}{4}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.44)+\sqrt{10.0736}}{2*2}=\frac{1.44+\sqrt{10.0736}}{4}
$