# How do I find all real and complex zeros of x^4 - 15x^2 - 75?

Finding the zeros of a bi-quadratic equation is rather easy if you know how to solve the quadratics. It's all about replacing ${x}^{2}$ by $y$.
Let $y = {x}^{2}$
 -> x^4−15x^2−75 $=$ ${y}^{2} - 15 y - 75$

Now you can use whatever method you like to find the zeros.
${y}_{1} = \frac{5}{2} \left(3 + \sqrt{21}\right)$ [positive]
${y}_{2} = \frac{5}{2} \left(3 - \sqrt{21}\right)$ [negative]//

For y=x²:
${x}_{1 , 1} = \sqrt{{y}_{1}}$ and ${x}_{1 , 2} = - \sqrt{{y}_{1}}$
${x}_{2 , 1} = \sqrt{{y}_{2}}$ and ${x}_{2 , 2} = - \sqrt{{y}_{2}}$

For ${y}_{1}$ is positive and ${y}_{2}$ is negative, ${x}_{1 , 1}$ and ${x}_{1 , 2}$ are real, but ${x}_{2 , 1}$ and ${x}_{2 , 2}$ are complex.

Hope it helps