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

Apr 29, 2015

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

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

Add the value of ${a}^{2}$ to both sides where
$2 a = - 6$
since we want ${\left(y + a\right)}^{2}$
$= {y}^{2} + 2 a x + {a}^{2}$
$= {y}^{2} - 6 y + {a}^{2}$

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

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

${\left(y - 3\right)}^{2} = 13$

$y - 3 = \pm \sqrt{13}$

$y = 3 \pm \sqrt{13}$