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Create a list of steps, in order, that will solve the following equation. 3(x+2)^(2)=48 Solution steps: Add 2 to both sides Divide both sides by 3 Multiply both sides by 3 Subtract 2 from both sides Square both sides Take the square root of both sides

Create a list of steps, in order, that will solve the following equation.\newline3(x+2)2=483(x+2)^{2}=48\newlineSolution steps:\newline- Add 22 to both sides\newline- Divide both sides by 33\newline- Multiply both sides by 33\newline- Subtract 22 from both sides\newline- Square both sides\newline- Take the square root of both sides

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Q. Create a list of steps, in order, that will solve the following equation.\newline3(x+2)2=483(x+2)^{2}=48\newlineSolution steps:\newline- Add 22 to both sides\newline- Divide both sides by 33\newline- Multiply both sides by 33\newline- Subtract 22 from both sides\newline- Square both sides\newline- Take the square root of both sides
  1. Divide and isolate squared term: Divide both sides by 33 to isolate the squared term.\newlineCalculation: 3(x+2)2=483(x+2)^2 = 48 becomes (x+2)2=483(x+2)^2 = \frac{48}{3}\newlineReasoning: To simplify the equation, we need to get rid of the coefficient in front of the squared term.\newlineMath error check:
  2. Simplify right side of equation: Simplify the right side of the equation.\newlineCalculation: (x+2)2=483(x+2)^2 = \frac{48}{3} becomes (x+2)2=16(x+2)^2 = 16\newlineReasoning: Dividing 4848 by 33 simplifies the equation further and prepares us to take the square root.\newlineMath error check:
  3. Take square root of both sides: Take the square root of both sides to solve for xx.\newlineCalculation: ((x+2)2)=16\sqrt{((x+2)^2)} = \sqrt{16} becomes x+2=±4x+2 = \pm 4\newlineReasoning: Taking the square root of both sides will help us find the value of xx.\newlineMath error check:
  4. Subtract to isolate x: Subtract 22 from both sides to isolate xx.\newlineCalculation: x+2=±4x+2 = \pm 4 becomes x=±4−2x = \pm 4 - 2\newlineReasoning: Subtracting 22 from both sides will give us the solution for xx.\newlineMath error check:
  5. Simplify equation for x: Simplify the equation to find the two possible values for x.\newlineCalculation: x=±4−2x = \pm 4 - 2 becomes x=4−2x = 4 - 2 or x=−4−2x = -4 - 2\newlineReasoning: Since we have a plus-minus situation, we need to consider both the positive and negative solutions.\newlineMath error check:
  6. Calculate final answers: Calculate the final answers.\newlineCalculation: x=4−2x = 4 - 2 becomes x=2x = 2 and x=−4−2x = -4 - 2 becomes x=−6x = -6\newlineReasoning: Performing the subtraction gives us the two solutions for xx.\newlineMath error check:

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