Solve for \( y \) \[ y^{2}+2 y-3=0 \]

To solve the equation \( y^{2} + 2y - 3 = 0 \), you can use the quadratic formula, which is given by: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, \( a = 1 \), \( b = 2 \), and \( c = -3 \). First, calculate the discriminant (\( b^2 - 4ac \)): \[ b^2 - 4ac = 2^2 - 4(1)(-3) = 4 + 12 = 16 \] Since the discriminant is positive, there will be two real solutions: \[ y = \frac{-2 \pm \sqrt{16}}{2(1)} = \frac{-2 \pm 4}{2} \] Calculating the two possible values for \( y \): 1. \( y = \frac{-2 + 4}{2} = \frac{2}{2} = 1 \) 2. \( y = \frac{-2 - 4}{2} = \frac{-6}{2} = -3 \) Thus, the solutions are \( y = 1 \) and \( y = -3 \).