Question

algebra ii - 1st block algebra 2 unit 2 final review 2025 solve the equation using the quadratic formula. ( y^2 - 4y + 29 = 0 ) the solution set is (\boxed{quad}). (simplify your answer. type an exact answer, using radicals as needed. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed. express complex numbers in terms of ( i ).)

Explanation:

Step1: Identify coefficients

For equation \(y^2 - 4y + 29 = 0\), \(a = 1\), \(b = -4\), \(c = 29\).

Step2: Apply quadratic formula

Quadratic formula: \(y=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Substitute values:
\(y=\frac{-(-4)\pm\sqrt{(-4)^2 - 4(1)(29)}}{2(1)}=\frac{4\pm\sqrt{16 - 116}}{2}\).
Simplify discriminant: \(\sqrt{-100}=\sqrt{100}i = 10i\).
Thus, \(y=\frac{4\pm10i}{2}=2\pm5i\).

Answer:

\(2 + 5i, 2 - 5i\)