-3(6v-9)9v=7(v+3)

Solution for -3(6v-9)9v=7(v+3) equation:

-3(6v-9)9v=7(v+3)

We move all terms to the left:

-3(6v-9)9v-(7(v+3))=0

We multiply parentheses

-162v^2+243v-(7(v+3))=0


We calculate terms in parentheses: -(7(v+3)), so:

7(v+3)

We multiply parentheses

7v+21

Back to the equation:

-(7v+21)


We get rid of parentheses

-162v^2+243v-7v-21=0

We add all the numbers together, and all the variables

-162v^2+236v-21=0

a = -162; b = 236; c = -21;
Δ = b2-4ac
Δ = 2362-4·(-162)·(-21)
Δ = 42088
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:

$\sqrt{\Delta}=\sqrt{42088}=\sqrt{4*10522}=\sqrt{4}*\sqrt{10522}=2\sqrt{10522}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(236)-2\sqrt{10522}}{2*-162}=\frac{-236-2\sqrt{10522}}{-324}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(236)+2\sqrt{10522}}{2*-162}=\frac{-236+2\sqrt{10522}}{-324}
$