1. Given equation

$3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0$

We need the number of real solutions.

Also note that $x\neq 0$ because of the terms $\frac1x$ and $\frac1{x^2}$.

2. Substitute a standard expression

Let

$t=x+\frac{1}{x}$

Then,

$x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=t^2-2$

Substitute into the equation:

$3(t^2-2)-2t+5=0$

$3t^2-6-2t+5=0$

$3t^2-2t-1=0$

3. Solve for $t$

$3t^2-2t-1=0$

Factorizing:

$3t^2-3t+t-1=0$

$3t(t-1)+1(t-1)=0$

$(3t+1)(t-1)=0$

So,

$t=1 \quad \text{or} \quad t=-\frac13$

That is,

$x+\frac1x=1 \quad \text{or} \quad x+\frac1x=-\frac13$

4. Check which values of $t$ are possible for real $x$

For real $x\neq 0$,

$x+\frac1x\ge 2 \quad \text{or} \quad x+\frac1x\le -2$

So the expression $x+\frac1x$ cannot take any value strictly between $-2$ and $2$.

But both values obtained are:

• $t=1$
• $t=-\frac13$

Both lie in the interval $(-2,2)$, so neither is possible for real $x$.

Hence there are no real solutions.

5. Conclusion

The number of real solutions is

$\boxed{0}$

So the correct option is C.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.