Answer
The solution set is $\emptyset$.
(no real solutions)
Work Step by Step
... $\quad 4^{x}=(2^{2})^{x}=(2^{x})^{2}$
... Substitute $t=2^{x}$ (t is positive)
... the equation becomes
$3t^{2}+4t+8=0 \quad $... use the quadratic formula
$t=\displaystyle \frac{-4\pm\sqrt{4^{2}-4(3)(8)}}{2(3)} \quad $... note: $4^{2}- 4(3)(8)=-80$
since the radicand (the discriminant) is negative, there are no real solutions
The solution set is $\emptyset$.
(no real solutions)
