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Use the quadratic formula to find the zeros of the function. y = 16x^2 + 40x + 25 A. {-1.25, 1.25} B. {-1.25} C. {-4, 5} D. {-0.8}
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\[ y = 16x^2 + 40x + 25 \\ x_{1,2} = \frac{-40 \pm \sqrt{40^2 - 4 \cdot 16 \cdot 25}}{2 \cdot 16} \\ x_{1,2} = \frac{-40 \pm \sqrt{1600 - 100 \cdot 16}}{32} \\ x_{1,2} = \frac{-40 \pm \sqrt{1600 - 1600}}{32} \\ x_{1,2} = \frac{-40 \pm 0}{32} \\ x = \frac{-40}{32} = -\frac{8 \cdot 5}{ 8 \cdot 4 } = -\frac{5}{4} = -1 \frac{1}{4} = -1.25 \]
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