To solve the equation \(2x^2 + 15 = -11x\), we first rearrange it to bring all terms to one side:
\[ 2x^2 + 11x + 15 = 0 \]
Next, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 11\), and \(c = 15\).
First, we calculate the discriminant \(b^2 - 4ac\):
\[ b^2 - 4ac = 11^2 - 4 \cdot 2 \cdot 15 = 121 - 120 = 1 \]
Now we can find the roots:
\[ x = \frac{-11 \pm \sqrt{1}}{2 \cdot 2} = \frac{-11 \pm 1}{4} \]
This will give us two solutions:
- \(x = \frac{-11 + 1}{4} = \frac{-10}{4} = -2.5\)
- \(x = \frac{-11 - 1}{4} = \frac{-12}{4} = -3\)
Thus, the solution set of the equation is:
\[ {-3, -2.5} \]
So the correct answer is:
D {-3, -2.5}.