The number of real solution of  $3(x^2+\frac{1}{x^2})-2(x+\frac{1}{x})+5=0$

The given equation is $3(x^2+\frac{1}{x^2})-2(x+\frac{1}{x})+5=0$

Suppose that $x+\frac{1}{x}=t$

It implies that 

 $\begin{array}{rcl}3\left(t^2-2\right)-2t+5& =& 0\\ 3t^2-2t-1& =& 0\\ \left(3t+1\right)\left(t-1\right)& =& 0\end{array}$

Hence, 

$t=-\frac{1}{3},1$

But the minimum value of $x+\frac{1}{x}$ is 2 when $x>0$ 

Hence, $x+\frac{1}{x}=1$ has no real solutions

Suppose that $x+\frac{1}{x}=-\frac{1}{3}$, it means $x$ is negative, but the maximum value of $x+\frac{1}{x}$ is $-2$

Hence, there is no real solution for the equation $x+\frac{1}{x}=-\frac{1}{3}$

Therefore, the number of real solutions for the equation $3(x^2+\frac{1}{x^2})-2(x+\frac{1}{x})+5=0$ is zero.