To solve the quadratic inequality \(-2x^2 + x + 15 \geq 0\), we first need to analyze the related quadratic equation:
\[ -2x^2 + x + 15 = 0 \]
We can apply the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = -2\), \(b = 1\), and \(c = 15\):
- Calculate the discriminant:
\[ b^2 - 4ac = 1^2 - 4(-2)(15) = 1 + 120 = 121 \]
- Since the discriminant is positive, there are two distinct real roots. Now, calculate the roots:
\[ x = \frac{-1 \pm \sqrt{121}}{2(-2)} = \frac{-1 \pm 11}{-4} \]
Calculating the two roots:
- For \(x = \frac{-1 + 11}{-4} = \frac{10}{-4} = -\frac{5}{2}\)
- For \(x = \frac{-1 - 11}{-4} = \frac{-12}{-4} = 3\)
So the roots are \(x = -\frac{5}{2}\) and \(x = 3\).
Next, we can determine the intervals where the quadratic \( -2x^2 + x + 15 \) is greater than or equal to zero. The roots divide the number line into three intervals:
- \( (-\infty, -\frac{5}{2}) \)
- \( \left(-\frac{5}{2}, 3\right) \)
- \( (3, \infty) \)
Now we will test a point from each interval to see where the quadratic is greater than or equal to zero.
-
Interval \( (-\infty, -\frac{5}{2}) \): Test \(x = -3\) \[ -2(-3)^2 + (-3) + 15 = -18 - 3 + 15 = -6 < 0 \]
-
Interval \( \left(-\frac{5}{2}, 3\right) \): Test \(x = 0\) \[ -2(0)^2 + (0) + 15 = 15 > 0 \]
-
Interval \( (3, \infty) \): Test \(x = 4\) \[ -2(4)^2 + (4) + 15 = -32 + 4 + 15 = -13 < 0 \]
From this analysis, we find that the quadratic is non-negative in the interval \( \left[-\frac{5}{2}, 3\right] \).
Including the roots where the quadratic equals zero, the final solution set in inequality notation is:
\[ \boxed{[-\frac{5}{2}, 3]} \]