Originally Posted by

**Soroban** Hello, Furyan!

As skeeter pointed out, this is a quadratic equation.

There is a way to recognize this phenomenon.

There are usually three terms, written in standard order.

If the first term has *twice* the exponent of the second term,

. . we may have a quadratic.

In your problem: .$\displaystyle e^{2(x)} - e^x - 12 \:=\:0 $

Another example: .$\displaystyle x^6 - 7x^3 - 8 \:=\:0$

Note that we have: .$\displaystyle \left(x^3\right)^2 - 7(x^3) - 8 \:=\:0$

Let $\displaystyle u \,=\,x^3$

Then we have: .$\displaystyle u^2 - 7u - 8 \:=\:0$

. . $\displaystyle (u-8)(u+1) \:=\:0 \quad\Rightarrow\quad u \:=\:8,\text{-}1$

Back-substitute: .$\displaystyle \begin{Bmatrix}x^3 \:=\:8 & \Rightarrow & x \:=\:2 \\ x^3 \:=\:\text{-}1 & \Rightarrow & x \:=\:\text{-}1 \end{Bmatrix}$

And there are four complex roots as well.