How do I solve for t?

Question

$t + 2 = \sqrt 12 t - 3$ $t+2=√12t−3$

p.s. 12t-3 is all under the root

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Answer 1 · 2018-06-27T07:46:00

$t = 1, 7$ $t=1,7$

$t + 2 = \sqrt{12 t - 3}$ $t+2=sqrt(12t-3)$

${(t + 2)}^{2} = {(\sqrt{12 t - 3})}^{2}$ $(t+2)^(color(red)2)=(sqrt(12t-3))^(color(red)2)$

$t^{2} + 4 t + 4 = 12 t - 3$ $t^2+4t+4=12t-3$

$t^{2} - 8 t + 7 = 0$ $t^2-8t+7=0$

$t = \frac{8 \pm \sqrt{64 - 28}}{2}$ $t=(8+-sqrt(64-28))/2$

$t = \frac{8 \pm 6}{2}$ $t=(8+-6)/2$

$t = 1, 7$ $t=1,7$