How do you solve 2x^2-7x+12=0 by completing the square?

1 Answer
Nallasivam V
Nov 16, 2016

x=7/4+-(isqrt(-47))/4

Explanation:

Given -

2x^2-7x+12=0

2x^2-7x=-12
Divide both sides by 2
(cancel(2)x^2)/cancel(2)-7/2x=(-12)/2
x^2-7/2x=-6
Divide 7/2 by 2, square it and add it to both sides

x^2-7/2*1/2*x +49/16=-6+49/16
x^2-7/4+49/16=-6+49/16
x^2-7/4+49/16=(-96+49)/16
(x-7/4)^2=(-47)/16
x-7/4=+-sqrt((-47)/16)
x=7/4+-(isqrt(-47))/4

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