How do you solve $y^2-6y-4=0$ by completing the square?

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Answer 1 · 2015-04-29T12:15:29

$y^2-6y-4 = 0$

Move the constant term to the right side:
$y^2-6y = 4$

Add the value of $a^2$ to both sides where
$2a = -6$
since we want $(y+a)^2$
$= y^2 +2ax +a^2$
$= y^2-6y+a^2$

$a =-3$
$a^2 = 9$

$y^2-6y+9= 4+9$

$(y-3)^2 = 13$

$y-3 = +-sqrt(13)$

$y = 3 +-sqrt(13)$

How do you solve y^2-6y-4=0y^2-6y-4=0 by completing the square?