3(v+2)2v=7(v-1)+9

Solution for 3(v+2)2v=7(v-1)+9 equation:

3(v+2)2v=7(v-1)+9

We move all terms to the left:

3(v+2)2v-(7(v-1)+9)=0

We multiply parentheses

6v^2+12v-(7(v-1)+9)=0


We calculate terms in parentheses: -(7(v-1)+9), so:

7(v-1)+9

We multiply parentheses

7v-7+9

We add all the numbers together, and all the variables

7v+2

Back to the equation:

-(7v+2)


We get rid of parentheses

6v^2+12v-7v-2=0

We add all the numbers together, and all the variables

6v^2+5v-2=0

a = 6; b = 5; c = -2;
Δ = b2-4ac
Δ = 52-4·6·(-2)
Δ = 73
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{-(5)-\sqrt{73}}{2*6}=\frac{-5-\sqrt{73}}{12}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*6}=\frac{-5+\sqrt{73}}{12}
$