How do you solve $5x^2-7=0$ ?

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Answer 1 · 2015-09-25T21:37:36

$x=+-sqrt(35)/5$

You can solve this by isolating $x$ .

$5x^2−7=0$
$5x^2−7+7=0+7$
$5x^2=7$
$1/5(5x^2)=1/5(7)$
$x^2=7/5$
$x=+-sqrt(7/5)$

You can have that as your final answer or you can rationalize it:
$x=+-sqrt(7/5)$
$x=+-sqrt(7/5*5/5)$
$x=+-sqrt((7*5)/5^2)$
$x=+-sqrt(35)/5$

How do you solve 5x^2-7=05x^2-7=0?