Proof: the following equation has only one real root

Hi. Could somebody show me how to prove that $ae^x=1+x+\frac{x^2}{2}$ has only one real root for $a>0$ ? All I know so far is that this equation has at least one root because $1+x+\frac{x^2}{2}>0$ for all real x.
I got a solution that i thought was cool so im posting anyways. Now if $ae^x = 1 + x + \frac{x^2}{2}$. then $y = ae^x$ and $y = 1 + x + \frac{x^2}{2}$ should be solutions to the differential eqn $y' - y = 0$ but for $y = 1 + x + \frac{x^2}{2}$, we get $y ' - y = \frac{x^2}{2} =0$ implies $x = 0$ so to have a solution to $ae^x = 1 + x + \frac{x^2}{2}$, x = 0 on the right side, so you get $ae^x = 1$ which for a > 1, has only 1 real root, namely $x = ln(\frac{1}{a})$