How do you solve $6x²-81=0$ ?

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Answer 1 · 2015-05-23T16:53:14

We can manipulate the expression by isolating $x$ . Alternatively, we can proceed to use Bhaskara.

Isolating $x$

$6x^2-81=0$
$x^2=81/6$
$x=sqrt(81/6)$
$x=+-9/sqrt(6)$
Rationalizing:
$x=+-(3sqrt(6))/6$

Using Bhaskara, where $a=6$ , $b=0$ and $c=-81$

$(-0+-sqrt(0-4(6)(-81)))/12$
$(+-sqrt(1944))/12$
$(+-sqrt(2^3*3^5))/12$
$(+-18sqrt(6))/12$

$+-(3sqrt(6))/2$

P.s.: there are other forms, but these seem so simple they appeal to me!

How do you solve 6x²-81=06x²-81=0?