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solve 3y^2-72=0
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\[3y^2-72=0\]
Sp, we're solving for y.
So*
subtract 3y^2 to both sides.
\(\bf 3y^2-72=0\implies 3y^2\cancel{ -72+72 }=0+72\implies 3y^2=72 \\ \quad \\ \cfrac{\cancel{ 3 }y^2}{\cancel{ 3 }}=\cfrac{\cancel{ 72 }}{\cancel{ 3 }}\implies y^2=24 \\ \quad \\ taking\qquad \sqrt{\qquad } \\ \quad \\ \sqrt{y^2}=\sqrt{24}\implies ?\)
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I can't think right now, but your answer would be y=2 sqrt6
@jdoe0001 will that be correct?
y = 2 square root 6
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