How do you solve #x^2=9# using the quadratic formula?

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Answer 1 · 2016-02-27T15:21:55

#x=+-3#

Write as :#" "x^2-9=0#

Compare to standard form of:#" "ax^2+bx+c=0#

Where:#" " x=(-b+-sqrt(b^2-4ac))/(2a)#

#a=1#
#b=0#
#c=-9#

So by substitution you have:

#x=(-0+-sqrt(0^2-4(1)(-9)))/(2xx1)#

#x=+-sqrt(+36)/2#

#x=+-sqrt(6^2)/2#

#x=+-3#

Tony B