Question

Extracting Square Roots.

2(t-3)² - 72 = 0​

Answer 1

Hope this helps you

Answer 2

$Given \; 2(t-3)^{2} - 72 = 0$

/* Dividing each term by 2, we get */

$\implies \frac{2(t-3)^{2} }{2} - \frac{72}{2} = 0$

$\implies (t-3)^{2} - 36 = 0$

$\implies (t-3)^{2} - 6^{2} = 0$

$\implies (t-3)^{2} = 6^{2}$

$\implies t-3 = \pm \sqrt{6^{2}}$

$\implies t-3 = \pm 6$

$\implies t-3 = -6 \:Or \: t-3 = 6$

$\implies t = -6+3 \:Or \: t = 6 + 3$

$\implies t = -3 \:Or \: t = 9$

Therefore.,

$\green { t = -3 \:Or \: t = 9}$

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