Question

\\3(x - 5)^2 + 2 = -16\\

Explanation:

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<reasoning>

Isolate the squared term

We begin by subtracting \(2\) from both sides of the equation to isolate the term containing the square.
\[
3(x - 5)^2 = -16 - 2
\]
\[
3(x - 5)^2 = -18
\]

Divide by the coefficient

Next, we divide both sides of the equation by \(3\) to fully isolate the squared binomial.
\[
(x - 5)^2 = \frac{-18}{3}
\]
\[
(x - 5)^2 = -6
\]

Take the square root

To solve for \(x\), we take the square root of both sides. Since the right side is negative, this introduces the imaginary unit \(i\), where \(i = \sqrt{-1}\).
\[
x - 5 = \pm\sqrt{-6}
\]
\[
x - 5 = \pm i\sqrt{6}
\]

Solve for x

Finally, we add \(5\) to both sides of the equation to find the complex solutions for \(x\).
\[
x = 5 \pm i\sqrt{6}
\]
</reasoning>

<answer>
\(x = 5 \pm i\sqrt{6}\)
</answer>

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Answer:

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<reasoning>

Isolate the squared term

We begin by subtracting \(2\) from both sides of the equation to isolate the term containing the square.
\[
3(x - 5)^2 = -16 - 2
\]
\[
3(x - 5)^2 = -18
\]

Divide by the coefficient

Next, we divide both sides of the equation by \(3\) to fully isolate the squared binomial.
\[
(x - 5)^2 = \frac{-18}{3}
\]
\[
(x - 5)^2 = -6
\]

Take the square root

To solve for \(x\), we take the square root of both sides. Since the right side is negative, this introduces the imaginary unit \(i\), where \(i = \sqrt{-1}\).
\[
x - 5 = \pm\sqrt{-6}
\]
\[
x - 5 = \pm i\sqrt{6}
\]

Solve for x

Finally, we add \(5\) to both sides of the equation to find the complex solutions for \(x\).
\[
x = 5 \pm i\sqrt{6}
\]
</reasoning>

<answer>
\(x = 5 \pm i\sqrt{6}\)
</answer>

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