find all real or imaginary solutions please any one know?

Question

3y^2+ 4v-1=0

Answer 1

This is a quadratic equation, assuming the 4v term is 4y, and the 4v was a typing error.

For solving quadratics, the formula is
y={-b+/-rt(b^2-4ac)} / 2a
a= coefficient of x^2 term, here+3
b= coefficient of x term, here +4
c =constant term, here -1
y={-4+/- rt(16-4(3)(-1))} /6
y={-4 +/- rt(16+12)} /6
y={-4 +/- rt(28)}/6
y={-4+/-2rt(7)}/6
y=2{-2+/-rt(7)}/6
y={-2+/-rt(7)}/3
y=[-2+rt(7)]/3 or [-2-rt(7)]/3

It is usual to leave the answers as shown. However, if you require a specific number for an answer, then
y=[-2+2.646]/3, = 0.215, or
y=[-2-2.646]/3, = -1.549

Answer 2

3y^2+ 4v-1=0

As written, the solution is:
y = +/- sqrt( (1-4v)/3 )
which is defined for all v except v=1/4.
y is real for real v<1/4,
pure imaginary for v>1/4,
undefined for v=1/4, and
complex for all other (complex) v.

If the v was supposed to be a y, then the quadratic formula gives:
y = (-4 +/- sqrt(4*4 - 4*3*(-1))) / (2*3)
= (-2/3) +/- sqrt(16+12)/6
= (-2/3) +/- sqrt(7)/3

Answer 3

I think you miss wrote a "v" instead of a "y"

The easiest way is to use the quadratic formula
(a*y^2 + b*y^2 + c) = (y + r1)(y + r2)

r1 = (-b + sqrt(b^2 -4*a*c))/2a
r2 = (-b - sqrt(b^2 -4*a*c))/2a

a = 3
b = 4
c = -1

r1 = (-4 + sqrt(4^2 - 4*3*(-1)))/(2*3) =
(-4 + sqrt(28))/6 = 0.21525

r2 = (-4 - sqrt(4^2 - 4*3*(-1)))/(2*3) =
(-4 - sqrt(28))/6 = -1.54858

answer check:
(y + 0.21525)(y - 1.54858) =
y^2 + 1.3333y -0.3333

note that the roots of 3y^2 + 4y - 1 = 0
divide both sides by 3
is the same as the roots of
y^2 +1.333y - 0.333 = 0

Answer 4

I don't think there is an answer since you only have one equation yet you have two variables (y and v) unless one of those are a typo.

Or are you supposed to find an answer in terms of one of those variables?

Answer 5

my imaginary solution involves goofy eating smoothies w/ george washington carver on the back of a rhinoceros.

Answer 6

hehe