How many real and imaginary roots are there for the equation 3n^2=-7n-7
I think did most of the work but I want somebody to check my answers

Question

anonymous · Answer

If you rearrange it into binomial form I think it should be:
3n^2-7n-7=0

anonymous · Answer

I think that the roots of this equation are:
$$n=\frac{7\pm \sqrt{133}}{6}$$
and
$$n \approx3.0888, -0.7554$$

anonymous · Answer

I need to know if I'm right so far and how many real and imaginary roots there are for the equation.

apoorvk · Answer

Is there any imaginary root? imaginary nos. include 'i' - iota.

i = sqrt(-1)

Now, none of your roots are in that form - so both are real!!

anonymous · Answer

Thank you so much.

apoorvk · Answer

Also, remember, that imaginary roots always appear in pairs - they are always in the form of conjugate surds!