To solve the equation
\[ \sqrt[3]{2x^2 + 19x + 17} = 2, \]
we'll start by eliminating the cube root by cubing both sides:
\[ 2x^2 + 19x + 17 = 2^3. \]
Calculating \(2^3\):
\[ 2^3 = 8. \]
Now we have the equation:
\[ 2x^2 + 19x + 17 = 8. \]
Next, we subtract 8 from both sides:
\[ 2x^2 + 19x + 17 - 8 = 0, \]
which simplifies to:
\[ 2x^2 + 19x + 9 = 0. \]
Now we can use the quadratic formula to find the solutions for \(x\). The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \(a = 2\), \(b = 19\), and \(c = 9\).
Calculating the discriminant (\(b^2 - 4ac\)):
\[ b^2 = 19^2 = 361, \] \[ 4ac = 4 \cdot 2 \cdot 9 = 72, \] \[ b^2 - 4ac = 361 - 72 = 289. \]
Now we take the square root of the discriminant:
\[ \sqrt{289} = 17. \]
Now we can plug this back into the quadratic formula:
\[ x = \frac{-19 \pm 17}{2 \cdot 2}. \]
This gives us two potential solutions:
- First solution:
\[ x = \frac{-19 + 17}{4} = \frac{-2}{4} = -0.5. \]
- Second solution:
\[ x = \frac{-19 - 17}{4} = \frac{-36}{4} = -9. \]
Therefore, the two solutions are
\[ x = -0.5 \quad \text{and} \quad x = -9. \]
Rounding to one decimal place if needed:
\[ x = -0.5, \quad -9. \]
So the final answers are:
\[ x = -0.5 \quad \text{and} \quad x = -9. \]