To solve the radical equation:
\[ \sqrt[3]{2x^2 + 19x + 17} = 2 \]
we begin by eliminating the cube root on the left side. We do this by cubing both sides of the equation:
\[ 2x^2 + 19x + 17 = 2^3 \]
Calculating \(2^3\):
\[ 2^3 = 8 \]
So the equation simplifies to:
\[ 2x^2 + 19x + 17 = 8 \]
Next, we subtract 8 from both sides to set the equation to zero:
\[ 2x^2 + 19x + 17 - 8 = 0 \]
This simplifies to:
\[ 2x^2 + 19x + 9 = 0 \]
Now, we can use the quadratic formula to solve for \(x\), where \(a = 2\), \(b = 19\), and \(c = 9\):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
First, we calculate the discriminant:
\[ b^2 - 4ac = 19^2 - 4 \cdot 2 \cdot 9 \]
Calculating \(19^2\):
\[ 19^2 = 361 \]
Calculating \(4 \cdot 2 \cdot 9\):
\[ 4 \cdot 2 \cdot 9 = 72 \]
Now subtract:
\[ 361 - 72 = 289 \]
The discriminant is \(289\). Now we can find the solutions using the quadratic formula:
\[ x = \frac{-19 \pm \sqrt{289}}{2 \cdot 2} \]
Since \( \sqrt{289} = 17 \), we have:
\[ x = \frac{-19 \pm 17}{4} \]
Calculating both potential solutions:
- First solution:
\[ x = \frac{-19 + 17}{4} = \frac{-2}{4} = -\frac{1}{2} = -0.5 \]
- Second solution:
\[ x = \frac{-19 - 17}{4} = \frac{-36}{4} = -9 \]
The two solutions are:
\[ x = -0.5 \quad \text{and} \quad x = -9 \]
Thus, the two solutions are:
\[ \boxed{-0.5} \quad \text{and} \quad \boxed{-9} \]