3(v-2)+5=2v(3v-5)

Solution for 3(v-2)+5=2v(3v-5) equation:

3(v-2)+5=2v(3v-5)

We move all terms to the left:

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

We multiply parentheses

3v-(2v(3v-5))-6+5=0


We calculate terms in parentheses: -(2v(3v-5)), so:

2v(3v-5)

We multiply parentheses

6v^2-10v

Back to the equation:

-(6v^2-10v)


We add all the numbers together, and all the variables

3v-(6v^2-10v)-1=0

We get rid of parentheses

-6v^2+3v+10v-1=0

We add all the numbers together, and all the variables

-6v^2+13v-1=0

a = -6; b = 13; c = -1;
Δ = b2-4ac
Δ = 132-4·(-6)·(-1)
Δ = 145
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{-(13)-\sqrt{145}}{2*-6}=\frac{-13-\sqrt{145}}{-12}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{145}}{2*-6}=\frac{-13+\sqrt{145}}{-12}
$