122. Solve the equation

$$
4 x^{4}+12 x^{3}+5 x^{2}-6 x-15=0
$$
Solution. We have:

$$
\begin{gathered}
4 x^{4}+12 x^{3}+5 x^{2}-6 x-15= \\
=\left(4 x^{4}+12 x^{3}+9 x^{2}\right)-\left(4 x^{2}+6 x+15\right)= \\
=\left(2 x^{2}+3 x\right)^{2}-2\left(2 x^{2}+3 x\right)-15= \\
=\left(2 x^{2}+3 x-5\right)\left(2 x^{2}+3 x+3\right) .
\end{gathered}
$$

Since the discriminant of the quadratic $2 x^{2}+3 x+3$ is negative, the set of solutions of the given equation coincides with the set of solutions of the equation $2 x^{2}+3 x-5=0$.

Answer: $x_{1}=1, x_{2}=-2.5$.