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

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

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Answer 1 · 2015-05-10T12:28:50

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

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

${(y - \frac{3}{2})}^{2} = \frac{65}{4}$ $(y - 3/2)^2 = 65/4$

Hence $y = \pm \sqrt{\frac{65}{4}} + \frac{3}{2} = \pm \frac{\sqrt{65}}{2} + \frac{3}{2}$ $y = +-sqrt(65/4)+3/2 = +-sqrt(65)/2+3/2$ .

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

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