(-2v2+4v)=(4v+5v2)

Solution for (-2v2+4v)=(4v+5v2) equation:

(-2v^2+4v)=(4v+5v^2)

We move all terms to the left:

(-2v^2+4v)-((4v+5v^2))=0

We get rid of parentheses

-2v^2-((4v+5v^2))+4v=0


We calculate terms in parentheses: -((4v+5v^2)), so:

(4v+5v^2)

We get rid of parentheses

5v^2+4v

Back to the equation:

-(5v^2+4v)


We add all the numbers together, and all the variables

-2v^2+4v-(5v^2+4v)=0

We get rid of parentheses

-2v^2-5v^2+4v-4v=0

We add all the numbers together, and all the variables

-7v^2=0

a = -7; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-7)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$v=\frac{-b}{2a}=\frac{0}{-14}=0$