How do you solve $9 r^{2} - 5 = 607$ $9r^2-5=607$ ?

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How do you solve $9 r^{2} - 5 = 607$ $9r^2-5=607$ ?

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Answer 1 · 2016-07-30T08:52:45

$r = \pm 2 \sqrt{17}$ $r=+-2sqrt17$

Explanation:

$9 r^{2} - 5 = 607$ $9r^2-5=607$
or
$9 r^{2} = 607 + 5$ $9r^2=607+5$
or
$9 r^{2} = 612$ $9r^2=612$
or
$r^{2} = \frac{612}{9}$ $r^2=612/9$
or
$r^{2} = 68$ $r^2=68$
or
$r = \pm \sqrt{68}$ $r=+-sqrt68$
or
$r = \pm \sqrt{(4) (17)}$ $r=+-sqrt((4)(17))$
or
$r = \pm 2 \sqrt{17}$ $r=+-2sqrt17$

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How do you solve $9 r^{2} - 5 = 607$ $9r^2-5=607$ ?

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How do you solve $9 r^{2} - 5 = 607$ $9r^2-5=607$ ?