288. Solve the equation:

$$
x^{4}+2 x^{3}-11 x^{2}+4 x+4=0
$$
$\triangle$ The given equation is not reciprocal. However, let's try to solve it using the same method as for a reciprocal equation, dividing the left part term by term by $x^{2}$:

$$
x^{2}+2 x-11+\frac{4}{x}+\frac{4}{x^{2}}=0, \quad\left(x^{2}+\frac{4}{x^{2}}\right)+2\left(x+\frac{2}{x}\right)-11=0
$$

Let $x+\frac{2}{x}=y$. Then

$$
x^{2}+4+\frac{4}{x^{2}}=y^{2}, \quad x^{2}+\frac{4}{x^{2}}=y^{2}-4 .
$$

We get:

$$
y^{2}-4+2 y-11=0, \quad y^{2}+2 y-15=0
$$

From the quadratic equation, we find $y$, and then from the substitution $-x$.

Answer: $1,2,(-5 \pm \sqrt{17}) / 2$.