Isolate term with t: First, we need to isolate the term containing t on one side of the equation. 5(t+2)² - 3 = 17 Add 3 to both sides to move the constant term to the right side of the equation. 5(t+2)² - 3 + 3 = 17 + 3 5(t+2)² = 20Add and simplify: Next, we divide both sides by 5 to solve for the squared term. 5(t+2)² / 5 = 20 / 5 (t+2)² = 4Divide by 5: Now, we take the square root of both sides to solve for t+2. √((t+2)²) = √4 t+2 = ±2Take square root: Finally, we subtract 2 from both sides to solve for t. t + 2 - 2 = ±2 - 2 t = 0 or t = -4

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(e) $5(t+2)^{2}-3=17$

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Algebra 2

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Q. (e)

5(t+2)^{2}-3=17

Isolate term with t: First, we need to isolate the term containing t on one side of the equation. $\newline$ $5(t+2)^2 - 3 = 17$ $\newline$ Add $3$ to both sides to move the constant term to the right side of the equation. $\newline$ $5(t+2)^2 - 3 + 3 = 17 + 3$ $\newline$ $5(t+2)^2 = 20$
Add and simplify: Next, we divide both sides by $5$ to solve for the squared term. $\frac{5(t+2)^2}{5} = \frac{20}{5}$ $(t+2)^2 = 4$
Divide by $5$ : Now, we take the square root of both sides to solve for $t+2$ . $\sqrt{(t+2)^2} = \sqrt{4}$ $t+2 = \pm 2$
Take square root: Finally, we subtract $2$ from both sides to solve for $t$ . $\newline$ $t + 2 - 2 = \pm 2 - 2$ $\newline$ $t = 0 \text{ or } t = -4$