1. Substitute to reduce the equation

Let $t=e^x.$ Since $e^x>0$ for all real $x$, we have $t>0.$

The given equation becomes $t^4+4t^3-58t^2+4t+1=0.$

2. Observe the symmetry

The polynomial is palindromic: $t^4+4t^3-58t^2+4t+1.$ So we divide by $t^2$ (valid since $t>0$): $t^2+4t-58+\frac{4}{t}+\frac{1}{t^2}=0.$

Rearrange: $\left(t^2+\frac{1}{t^2}\right)+4\left(t+\frac{1}{t}\right)-58=0.$

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

Let $y=t+\frac{1}{t}.$ Then the equation becomes $y^2-2+4y-58=0,$ so $y^2+4y-60=0.$

3. Solve for $y$

$y^2+4y-60=0$ $\Rightarrow (y+10)(y-6)=0.$ Thus, $y=6 \quad \text{or} \quad y=-10.$

4. Use the constraint on $y$

Since $t>0$, $t+\frac{1}{t}\ge 2.$ Hence $y\ge 2$.

So $y=-10$ is impossible, and only $y=6$ is valid.

5. Solve for $t$

We now solve $t+\frac{1}{t}=6.$ Multiplying by $t$: $t^2-6t+1=0.$

Thus, $t=\frac{6\pm\sqrt{36-4}}{2}=\frac{6\pm\sqrt{32}}{2}=3\pm 2\sqrt{2}.$

Both values are positive, so both are valid values of $t=e^x$.

6. Find the number of real $x$

Since $e^x=t$, each positive $t$ gives exactly one real $x=\ln t$.

Therefore there are $2$ real solutions.

Comparison with stored answer: Stored correct answer is $2$, which matches the derived answer.