Answer
$x_1=3$ and $x_2=-1$
Work Step by Step
$e^{x^2-3}=e^{2x}$
Applying the One-to-One Property:
$x^2-3=2x$
$x^2-2x-3=0$ ($a=1,~b=-2,~c=-3$)
$x=\frac{-(-2)±\sqrt {(-2)^2-4(1)(-3)}}{2(1)}$
$x=\frac{2±\sqrt {16}}{2}=\frac{2±4}{2}$
$x_1=\frac{6}{2}=3$ and $x_2=\frac{-2}{2}=-1$
