quadratic equation with imaginary part How can we  solve this equation 

$x^2-bx-c=0$ where c is not real number.

I tried to solve the above equation but I am not sure if it is correct or not.

$x_{1,2}=\frac{b}{2}\pm \frac{\sqrt{b^2+4c}}{2}$

I do not know what is the next step to solve this problem when  c is not real number.
The method is exactly the same as if $c$ was real. Whatever $c$ is, just leave it, until you have your answer in the $x = \ldots$ format. At that point, you may expand/factor the expression as you see fit.

Note that $x_{1,2}$ will likely end up being complex solutions. So getting them on the cartesian form $x = a + bi$ might be handy if you need to use the solutions further on.

As for $\sqrt{b^2 + 4c}$, we have a complex in there (which I failed to address, pre-edit - sorry!), which means that the entire expression is complex. We may or may not find that the complex number has a zero imaginary part, in which case it's real, but we'll work with it as a complex number of the form $r + si$.

$$\sqrt{b^2 + 4c} = r+si$$

Square both sides, and expand RHS.

$$b^2 + 4c = r^2 - s^2 + 2rsi$$

With $c = p + qi$ being complex, we can rewrite LHS, and have:

$$b^2 + 4(p + qi) = r^2 - s^2 + 2rsi$$

$$b^2 + 4p + 4qi = r^2 - s^2 + 2rsi$$

Each side has a real part and an imaginary part. For the equality to hold, each of those must equal eachother. I.e.

$$b^2 + 4p = r^2 - s^2 \ \wedge 4qi = 2rsi$$

From there, you can rewrite the expression under the square root sign while taking into account that $c$ is complex.

Note; $p, q, r, s$ are all real.
The method is exactly the same as if $c$ was real. Whatever $c$ is, just leave it, until you have your answer in the $x = \ldots$ format. At that point, you may expand/factor the expression as you see fit.

Note that $x_{1,2}$ will likely end up being complex solutions. So getting them on the cartesian form $x = a + bi$ might be handy if you need to use the solutions further on.

As for $\sqrt{b^2 + 4c}$, we have a complex in there (which I failed to address, pre-edit - sorry!), which means that the entire expression is complex. We may or may not find that the complex number has a zero imaginary part, in which case it's real, but we'll work with it as a complex number of the form $r + si$.

$$\sqrt{b^2 + 4c} = r+si$$

Square both sides, and expand RHS.

$$b^2 + 4c = r^2 - s^2 + 2rsi$$

With $c = p + qi$ being complex, we can rewrite LHS, and have:

$$b^2 + 4(p + qi) = r^2 - s^2 + 2rsi$$

$$b^2 + 4p + 4qi = r^2 - s^2 + 2rsi$$

Each side has a real part and an imaginary part. For the equality to hold, each of those must equal eachother. I.e.

$$b^2 + 4p = r^2 - s^2 \ \wedge 4qi = 2rsi$$

From there, you can rewrite the expression under the square root sign while taking into account that $c$ is complex.

Note; $p, q, r, s$ are all real.