Solve using the quadratic formula. \( y^{2}+3 y+1=0 \) Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth. \( y=\square \) or \( y=\square \) Submit

To solve the quadratic equation \( y^2 + 3y + 1 = 0 \) using the quadratic formula, we use the formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this equation, \( a = 1 \), \( b = 3 \), and \( c = 1 \). First, we calculate the discriminant \( b^2 - 4ac \): \[ b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 1 = 9 - 4 = 5 \] Now we can substitute \( a \), \( b \), and the discriminant into the quadratic formula: \[ y = \frac{-3 \pm \sqrt{5}}{2 \cdot 1} = \frac{-3 \pm \sqrt{5}}{2} \] This gives us two solutions: 1. \( y = \frac{-3 + \sqrt{5}}{2} \) 2. \( y = \frac{-3 - \sqrt{5}}{2} \) Now, we can approximate these values to the nearest hundredth. Calculating \( \sqrt{5} \) (approximately 2.236): 1. For \( y = \frac{-3 + \sqrt{5}}{2} \): \[ y \approx \frac{-3 + 2.236}{2} = \frac{-0.764}{2} \approx -0.382 \] 2. For \( y = \frac{-3 - \sqrt{5}}{2} \): \[ y \approx \frac{-3 - 2.236}{2} = \frac{-5.236}{2} \approx -2.618 \] Thus, the solutions rounded to the nearest hundredth are: \[ y \approx -0.38 \quad \text{and} \quad y \approx -2.62 \] So, the final answers are: \[ y = \frac{-3 + \sqrt{5}}{2} \quad \text{or} \quad y = \frac{-3 - \sqrt{5}}{2} \] And in decimal form: \[ y \approx -0.38 \quad \text{or} \quad y \approx -2.62 \]