The number of real roots of the equation $$e^{4x} + 2e^{3x} - e^x - 6 = 0$$ is:

We have to solve the exponential equation

$$e^{4x}+2e^{3x}-e^{x}-6=0.$$

Because every exponential term contains the same base $$\,e\,$$ raised to a multiple of $$x$$, it is natural to set

$$t=e^{x}\qquad\bigl(\text{so }t\gt 0\bigr).$$

Substituting $$\,e^{x}=t\,$$ into each term gives

$$\begin{aligned} e^{4x}&=(e^{x})^{4}=t^{4},\\[2mm] 2e^{3x}&=2(e^{x})^{3}=2t^{3},\\[2mm] -e^{x}&=-t. \end{aligned}$$

Hence the original equation becomes

$$t^{4}+2t^{3}-t-6=0.$$

Now we try to factor this quartic. A convenient way is to look for a product of two quadratic factors

$$t^{4}+2t^{3}-t-6 =(t^{2}+at+b)(t^{2}+ct+d).$$

Expanding the right-hand side, we obtain

$$t^{4}+(a+c)t^{3}+(ac+b+d)t^{2}+(ad+bc)t+bd.$$

Comparing coefficients with $$\,t^{4}+2t^{3}-t-6\,$$ we get the system

$$\begin{cases} a+c=2,\\ ac+b+d=0,\\ ad+bc=-1,\\ bd=-6. \end{cases}$$

A little trial shows that choosing

$$b=2,\;d=-3,\;a=1,\;c=1$$

satisfies all four conditions, because

$$\begin{aligned} a+c&=1+1=2,\\[2mm] ac+b+d&=(1)(1)+2+(-3)=0,\\[2mm] ad+bc&=(1)(-3)+(2)(1)=-3+2=-1,\\[2mm] bd&=(2)(-3)=-6. \end{aligned}$$

Therefore

$$t^{4}+2t^{3}-t-6=(t^{2}+t+2)(t^{2}+t-3)=0.$$

So we must solve

$$t^{2}+t+2=0\quad\text{or}\quad t^{2}+t-3=0.$$

First consider $$\,t^{2}+t+2=0.$$ Its discriminant is

$$\Delta_1=1^{2}-4(1)(2)=1-8=-7\lt 0,$$

so this quadratic has no real solutions. Because $$\,t=e^{x}\,,$$ which is always positive and real, we discard this factor entirely.

Next solve $$\,t^{2}+t-3=0.$$ The discriminant here is

$$\Delta_2=1^{2}-4(1)(-3)=1+12=13\gt 0,$$

giving two real roots

$$t=\frac{-1\pm\sqrt{13}}{2}.$$

Numerically,

$$\frac{-1+\sqrt{13}}{2}\approx\frac{-1+3.605}{2}\approx1.3027\gt 0,$$

while

$$\frac{-1-\sqrt{13}}{2}\approx\frac{-1-3.605}{2}\approx-2.3027\lt 0.$$

The second value is negative and therefore impossible, because $$\,t=e^{x}\gt 0\,$$ for every real $$x$$. Only the positive root survives:

$$t=\frac{-1+\sqrt{13}}{2}.$$

Finally revert to $$\,t=e^{x}.$$ We obtain

$$e^{x}=\frac{-1+\sqrt{13}}{2}\;\;\Longrightarrow\;\;x=\ln\!\left(\frac{-1+\sqrt{13}}{2}\right).$$

This is a single real value of $$x$$. Therefore the given equation possesses exactly one real root.

So, the answer is $$1$$.