How do you solve $3x^2=48$ ?

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Answer 1 · 2018-03-04T18:29:25

$x=+-4$

We begin by dividing both sides by $3$ :

$(cancel3x^2)/cancel3=48/3$

$x^2=16$

Take the square root on both sides (and don't forget the negative root):

$sqrt(x^2)=+-sqrt(16)$

$x=+-4$

How do you solve 3x^2=48 3x^2=48?