How do you solve $x^2-7x+8=0$ by completing the square?

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Answer 1 · 2015-05-19T16:23:47

$y = x^2 - 7x + 8 = x^2 - 7x + 49/4 - 49/4 + 8 = 0$

$y = (x - 7/2)^2 - 49/4 + 32/4$

$(x - 7/2)^2 = 17/4$

$x = 7/2 + (sqrt17)/2$

$x = 7/2 - (sqrt17)/2$

How do you solve x^2-7x+8=0x^2-7x+8=0 by completing the square?