How do you solve $y^2 - 3y - 14 = 0$ by completing the square?

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Answer 1 · 2015-05-10T12:28:50

$(y-(3/2))^2 = y^2 - 3y + 9/4$ ,
so $y^2 - 3y - 14 = (y - 3/2)^2 - (14+9/4)$
$= (y-3/2)^2 - 65/4$ .

And, since we are given that $y^2 - 3y - 14 =0$

$(y - 3/2)^2 = 65/4$

Hence $y = +-sqrt(65/4)+3/2 = +-sqrt(65)/2+3/2$ .

How do you solve y^2 - 3y - 14 = 0 y^2 - 3y - 14 = 0 by completing the square?