How do you solve $x^{2} + 8 = 28$ $x^2+8=28$ ?

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Answer 1 · 2017-02-17T05:35:48

$x = \pm 4 \sqrt{5}$ $x=+-4sqrt5$

$x^{2} + 8 = 28$ $x^2+8=28$
$x^{2} = 20$ $x^2=20$
$\sqrt{x^{2}} = \sqrt{20}$ $sqrt(x^2)=sqrt(20)$

$x = \pm \sqrt{20} = \pm 4 \sqrt{5}$ $x=+-sqrt(20)=+-4sqrt5$

How do you solve x^2+8=28x2+8=28x^2+8=28?