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Solve for (x+5)^2=12
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x=-5+or- 2 radical 3
Suppose \[y^2 = 12\] What would say is the solution or are the solutions for y?
if only we had a sqrt(12) in there but that +5 is getting in the way .... i say we put a hit out on it :) -5 +- sqrt(12)
could it be -5-2√6,-5+2√6
almost; but no
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2sqrt(6) = sqrt(24)
-5-2√3,-5+2√3
\[\sqrt{12} = \sqrt{4 \times 3} = \sqrt{2^2 \times 3} = \sqrt{2^2} \times \sqrt{3} = 2 \sqrt{3}\]
-2√3 2√3 has to be close
No, you were right above\[x = -5 \pm 2 \sqrt{3}\]
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