# Question #c3298

Jun 21, 2015

This equation has no solutions.

#### Explanation:

First we can substitute $t = {e}^{x}$, this changes an exponential equation to a quadratic equation:

${t}^{2} + 8 t + 12 = 0$, but we only look for roots greater than zero, because ${e}^{x}$ cannot be a negative number.

If we calculate $\Delta$ we find, that it is positive:

$\Delta = {8}^{2} - 4 \cdot 1 \cdot 12 = 64 - 48 = 16$

$\sqrt{\Delta} = 4$

${t}_{1} = \frac{- 8 - 4}{2} = - 6 < 0$

${t}_{2} = \frac{- 8 + 4}{2} = - 2 < 0$

The quadratc equation has 2 solutions, but neither of them is positive, so the base exponential equation has no solutions.