How do you solve $2x^2-9=0$ using the quadratic formula?

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Answer 1 · 2016-05-11T02:45:31

$x=sqrt72/4$
$x=-(sqrt72/4)$

Given -

$2x^2-9=0$

Quadratic equations normally looks like this

$ax^2+bx+c=0$

In the given equation the $bx$ term is missing

We shall supply it

$2x^2+0x-9$

Then as per formula the roots are

$x=((-b)+-sqrt(b^2-4ac))/(2a)$

We shall substitute the values in the formula

$x=((-0)+-sqrt(0^2-(4xx2xx(-9))))/(2xx2)$

$x=(+-sqrt(0-(-72)))/4$

The two roots are

$x=sqrt72/4$
$x=-(sqrt72/4)$

How do you solve 2x^2-9=02x^2-9=0 using the quadratic formula?