v2=v02+2v(v+v0)

Solution for v2=v02+2v(v+v0) equation:

v2=v02+2v(v+v0)

We move all terms to the left:

v2-(v02+2v(v+v0))=0

We add all the numbers together, and all the variables

v2-(v02+2v(+2v))=0

We add all the numbers together, and all the variables

v^2-(v02+2v(+2v))=0


We calculate terms in parentheses: -(v02+2v(+2v)), so:

v02+2v(+2v)

We add all the numbers together, and all the variables

v^02+2v(+2v)

We multiply parentheses

v^02+4v^2

Back to the equation:

-(v^02+4v^2)


We get rid of parentheses

v^2-v^02-4v^2=0

We add all the numbers together, and all the variables

-3v^2-1v^02=0

a = -1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-1)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$v=\frac{-b}{2a}=\frac{0}{-2}=0$