Let's go through the steps Eliza might have taken to solve the quadratic equation in detail.
We start with the original equation:
[tex]\[ \frac{1}{5}(x+3)^2 - 2 = 18 \][/tex]
For Step 1, we need to isolate the square term. First, we add 2 to both sides:
[tex]\[ \frac{1}{5}(x+3)^2 - 2 + 2 = 18 + 2 \][/tex]
[tex]\[ \frac{1}{5}(x+3)^2 = 20 \][/tex]
Next, we multiply both sides of the equation by 5 to eliminate the fraction:
[tex]\[ 5 \cdot \frac{1}{5}(x+3)^2 = 5 \cdot 20 \][/tex]
[tex]\[ (x+3)^2 = 100 \][/tex]
This brings us to Step 2. Therefore, Eliza should have written:
[tex]\[ (x+3)^2 = 100 \][/tex]
Step 3 involves solving for [tex]\( x+3 \)[/tex] by taking the square root of both sides:
[tex]\[ x+3 = \pm \sqrt{100} \][/tex]
[tex]\[ x+3 = \pm 10 \][/tex]
Step 4 is to solve for [tex]\( x \)[/tex]:
[tex]\[ x+3 = 10 \quad \Rightarrow \quad x = 10 - 3 = 7 \][/tex]
[tex]\[ x+3 = -10 \quad \Rightarrow \quad x = -10 - 3 = -13 \][/tex]
Thus, the solutions are:
[tex]\[ x = -13 \text{ or } x = 7 \][/tex]
So, in summary, the result from Step 2 should be:
[tex]\[ (x+3)^2 = 100 \][/tex]
