Answer
There are no real solutions.
Work Step by Step
Re-write as: $49^x=(7^{2})^{x}=(7^{x})^{2}$
We wish to substitute $a=7^{x}$, so the equation becomes:
$2a^{2}+11a+5=0 $
This gives a quadratic equation, whose factors are: $(2a+1)(a+5)=0$
Use the zero factor property to obtain:
$ 2u+1 =0 \implies a=-\dfrac{1}{2}$
and
$a+5 =0 \implies a=-5$
Further, we will plug $a=7^x$ into the above value of $a$ to get the value of $x$ as follows: $7^x=\dfrac{-1}{2} $ and $7^x =-5$.
Since, $7^x \gt 0$, the solutions of $7^x=\dfrac{-1}{2} $ and $7^x =-5$ cannot be accepted.
Thus, there are no real solutions.
