To solve the quadratic equation \(15y^2 + 32y + 16 = 0\), we can use the quadratic formula, which is given by:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In your equation, \(a = 15\), \(b = 32\), and \(c = 16\).
Now we need to calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 = 32^2 = 1024 \] \[ 4ac = 4 \cdot 15 \cdot 16 = 960 \]
So, the discriminant is:
\[ b^2 - 4ac = 1024 - 960 = 64 \]
Now that we have the discriminant, we can find \(y\):
\[ y = \frac{-32 \pm \sqrt{64}}{2 \cdot 15} \]
The square root of 64 is 8, so we have:
\[ y = \frac{-32 \pm 8}{30} \]
This gives us two solutions:
- \(y = \frac{-32 + 8}{30} = \frac{-24}{30} = -\frac{4}{5}\)
- \(y = \frac{-32 - 8}{30} = \frac{-40}{30} = -\frac{4}{3}\)
So the solutions are:
\[ y = -\frac{4}{3}, \quad y = -\frac{4}{5} \]
The correct answer is B. \( -\frac{4}{3}, -\frac{4}{5} \).