Answer
The solution set is $\emptyset$.
(no real solutions)
Work Step by Step
... $\quad 49^{x}=(7^{2})^{x}=(7^{x})^{2}$
... Substitute $t=7^{x}$ (t is positive)
... the equation becomes
$2t^{2}+11t+5=0 \quad $... use the quadratic formula
$t=\displaystyle \frac{-11\pm\sqrt{11^{2}-4(2)(5)}}{2(2)} \quad $
$t==\displaystyle \frac{-11\pm\sqrt{81}}{4}$
$t=\displaystyle \frac{-11-9}{4}=-5$ or $t=\displaystyle \frac{-11+9}{4}=\frac{1}{2}.$
Neither t is acceptable, as $t=7^{x}$ cannot be negative.
The solution set is $\emptyset$.
(no real solutions)