5(v+3)=2(21+3v)2v

Solution for 5(v+3)=2(21+3v)2v equation:

5(v+3)=2(21+3v)2v

We move all terms to the left:

5(v+3)-(2(21+3v)2v)=0

We add all the numbers together, and all the variables

5(v+3)-(2(3v+21)2v)=0

We multiply parentheses

5v-(2(3v+21)2v)+15=0


We calculate terms in parentheses: -(2(3v+21)2v), so:

2(3v+21)2v

We multiply parentheses

12v^2+84v

Back to the equation:

-(12v^2+84v)


We get rid of parentheses

-12v^2+5v-84v+15=0

We add all the numbers together, and all the variables

-12v^2-79v+15=0

a = -12; b = -79; c = +15;
Δ = b2-4ac
Δ = -792-4·(-12)·15
Δ = 6961
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{-(-79)-\sqrt{6961}}{2*-12}=\frac{79-\sqrt{6961}}{-24}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-79)+\sqrt{6961}}{2*-12}=\frac{79+\sqrt{6961}}{-24}
$