Answer
No real solutions.
Work Step by Step
Re-write first term as:
$4^x=(2^{2})^{x}=(2^{x})^{2}$
We wish to substitute $a=2^{x}$, so the equation becomes:
$3a^{2}+4a+8=0 $
This gives a quadratic equation, whose factors can be found by using the quadratic formula.
$a=\dfrac{−4 \pm \sqrt {4^2-(4)(3)(8)}}{(2)(3)} = 42−4(3)(8)=−80$
We see the discriminant is negative, so there are no real solutions.
