How to find the x intercepts 
$\frac{4}{3} e^{3x} + 2 e^{2x} - 8 e^x$

I have some confusion especially because of the e     
how can I approach the solution?
The solution of the x-intercept is 0.838
Many thanks

\begin{align}
0&=\frac{4}{3}e^{3x}+2e^{2x}-8e^x\tag{1} \\[1em]
& = \frac{\frac{4}{3}e^{3x}+2e^{2x}-8e^{x}}{2e^x}\tag{2} \\[1em]
& = \frac{2}{3}e^{2x}+e^x-4\tag{3} \\[1em]
\end{align}
Now let $\xi=e^x,\therefore e^{2x}=\left(e^x\right)^2=\xi^2.$ This gives us
\begin{align}
0&=\frac{2}{3}\xi^2 +\xi-4\tag{4} \\[1em]
\therefore \xi & = \left\{\frac{-1+\sqrt{1-4\left(\frac{2}{3}\right)\left(-4\right)}}{2},\:\frac{-1-\sqrt{1-4\left(\frac{2}{3}\right)\left(-4\right)}}{2}\right\}\tag{5} \\[1em]
\end{align}
And I'm sure you can do the rest...
