How do you solve x^2-5x-3/4=0 by completing the square?

1 Answer
George C.
May 13, 2016

x = 5/2+-sqrt(7)

Explanation:

Premultiply by 4 first to cut down on the fractions involved:

0 = 4(x^2-5x-3/4)

=4x^2-20x-3

=(2x)^2-2(2x)(5)-3

=(2x-5)^2-5^2-3

=(2x-5)^2-28

=(2x-5)^2-2^2*7

=(2x-5)^2-(2sqrt(7))^2

=((2x-5)-2sqrt(7))((2x-5)+2sqrt(7))

=(2x-5-2sqrt(7))(2x-5+2sqrt(7))

=4(x-5/2-sqrt(7))(x-5/2+sqrt(7))

So x = 5/2+-sqrt(7)

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