# Explanation: ## Step1: Complete the square for the quadratic equation For the equation \(y^{2}+6y - 16=0\), we first focus on the \(y^{2}+6y\) part. To complete the square, we take half of the coefficient of \(y\), which is \(\frac{6}{2} = 3\), and then square it: \(3^{2}=9\). We add and subtract this value in the equation: \[ \begin{align*} y^{2}+6y + 9-9 - 16&=0\\ (y + 3)^{2}-25&=0 \end{align*} \] ## Step2: Solve for \(y\) We can rewrite the equation as \((y + 3)^{2}=25\). Taking the square root of both sides, we get \(y + 3=\pm5\). - Case 1: When \(y + 3 = 5\), then \(y=5 - 3=2\). - Case 2: When \(y + 3=-5\), then \(y=-5 - 3=-8\). # Answer: The solutions are \(y = 2\) and \(y=-8\)

y² + 6y - 16 = 0

QUESTION IMAGE

Question

y² + 6y - 16 = 0

Explanation:

Step1: Complete the square for the quadratic equation

For the equation \(y^{2}+6y - 16=0\), we first focus on the \(y^{2}+6y\) part. To complete the square, we take half of the coefficient of \(y\), which is \(\frac{6}{2} = 3\), and then square it: \(3^{2}=9\). We add and subtract this value in the equation:

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Step2: Solve for \(y\)

We can rewrite the equation as \((y + 3)^{2}=25\). Taking the square root of both sides, we get \(y + 3=\pm5\).

Case 1: When \(y + 3 = 5\), then \(y=5 - 3=2\).
Case 2: When \(y + 3=-5\), then \(y=-5 - 3=-8\).

Answer:

The solutions are \(y = 2\) and \(y=-8\)