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)#