How would you find $x$ in:
$e^{x^2+4x-7}(6x^2+12x+3)=0$
I don't know where to begin. Can you do the following?
$e^{x^2+4x-7}=1/(6x^2+12x+3)$
and then find $ln$ for both sides?
2026-04-04 05:15:45.1775279745
Solving exponential equation $e^{x^2+4x-7}(6x^2+12x+3)=0$
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$e^{x^2+4x-7}(6x^2+12x+3)=0 \Rightarrow e^{x^2+4x-7}=0 \text{ or } \ 6x^2+12x+3=0$
$$\text{It is known that } e^{x^2+4x-7} \text{ is non-zero }$$
therefore,you have to solve :
$$6x^2+12x+3=0$$
The solutions are:
$$x=-1-\frac{1}{\sqrt{2}} \\ x=-1+\frac{1}{\sqrt{2}}$$