Just a simple complex equation.

Solving the quadratic in $z^5$ we would get two roots each of which will yield five roots each which would together form the 10 roots of the original polynomial.

So we have $\displaystyle z^{5} = 32$ or $z^{5} = -31$

For $z^{5} = 32$

we have $\displaystyle z = 2e^{\frac{2ik\pi}{5}}$ where $k \in A$

$\displaystyle A = \{0,1,2,3,4\}$ .

Note that $A$ can be any set of five consecutive integers. For simplicity we have chosen $A$ to be $\displaystyle \{0,1,2,3,4\}$

Now the real part of the complex numbers would be negative for $k= \{2,3\}$ . So we have $2$ such roots.

For $z^{5} = -31$

we have $\displaystyle z= 31^{\frac{1}{5}} (-1)^{\frac{1}{5}}$

So $\displaystyle z=31^{\frac{1}{5}} e^{\frac{i\pi(2k+1)}{5}}$ where $k\in B$

$\displaystyle B = \{0,1,2,3,4\}$ .

Note again that $B$ can be any set of five consecutive integers . For simplicity we have chosen $\displaystyle B$ to be $\displaystyle \{0,1,2,3,4\}$ .

Here the real part of the complex numbers would be negative for $\displaystyle k= (1,2,3)$ . Which gives us $3$ more of such roots.

Hence in total there are $\displaystyle 5$ such complex numbers whose real parts are negative.