How do you solve $x^2+5=167$ ?

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Answer 1 · 2017-02-05T13:23:32

$x=+-9*sqrt2$ or $x=+-12.728$ to the nearest 3 decimals

$x^2+5=167" "$ there is no $x$ term change to $x^2 = c$

$:.x^2=162$

$:.x=+-sqrt162$

$:.x=+-(sqrt3*sqrt3*sqrt3*sqrt3*sqrt2)$

$sqrt3*sqrt3=3$

$:.x=+-(3*3*sqrt2)$

$:.x=+-9sqrt2$

Or $x=+-9*1.414213562$

$:.x=+-12.728$ to the nearest 3 decimals.

How do you solve x^2+5=167x^2+5=167?