3v(v+1)=(2-v)*2

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

3v(v+1)=(2-v)*2

We move all terms to the left:

3v(v+1)-((2-v)*2)=0

We add all the numbers together, and all the variables

3v(v+1)-((-1v+2)*2)=0

We multiply parentheses

3v^2+3v-((-1v+2)*2)=0


We calculate terms in parentheses: -((-1v+2)*2), so:

(-1v+2)*2

We multiply parentheses

-2v+4

Back to the equation:

-(-2v+4)


We get rid of parentheses

3v^2+3v+2v-4=0

We add all the numbers together, and all the variables

3v^2+5v-4=0

a = 3; b = 5; c = -4;
Δ = b2-4ac
Δ = 52-4·3·(-4)
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*3}=\frac{-5-\sqrt{73}}{6}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*3}=\frac{-5+\sqrt{73}}{6}
$