How do I solve for t?

Question

#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+2=sqrt(12t-3)#

#(t+2)^(color(red)2)=(sqrt(12t-3))^(color(red)2)#

#t^2+4t+4=12t-3#

#t^2-8t+7=0#

#t=(8+-sqrt(64-28))/2#

#t=(8+-6)/2#

#t=1,7#