# How do you solve n^2 + 19n + 66 = 6 by completing the square?

Apr 29, 2015

First simplify the process by moving the constant term to the right side of the equation;
then add whatever constant is necessary to both sides so that the left side is a square;
finally take the square root of both sides and simplify.

${n}^{2} + 19 n + 66 = 6$

${n}^{2} + 19 n = - 60$

If ${n}^{2} + 19 n$ are the first two terms of a square
${\left(n + a\right)}^{2} = \left({n}^{2} + 2 a n + {a}^{2}\right)$
then
$a = \frac{19}{2}$
and
${a}^{2} = {\left(\frac{19}{2}\right)}^{2} = \frac{361}{4} = 90 \frac{1}{4}$

${n}^{2} + 19 n + {\left(\frac{19}{2}\right)}^{2} = - 60 + 90 \frac{1}{4}$

${\left(n + \frac{19}{2}\right)}^{2} = 30 \frac{1}{4} = \frac{121}{4}$

$n + \frac{19}{2} = \pm \sqrt{\frac{121}{4}} = \pm \frac{11}{2}$

$n = - \frac{19}{2} \pm \frac{11}{2} = \frac{- 19 \pm 11}{2}$

$n = - 15$
or
$n = - - 4$

Apr 29, 2015

$n = - 4 , - 15$

You can solve the problem by factoring.

${n}^{2} + 19 n + 66 = 6$

Subtract 6 from both sides.

${n}^{2} + 19 n + 60 = 0$

Factor.

$4 \times 15 = 60 \mathmr{and} 4 + 15 = 19$

$\left(n + 4\right) \left(n + 15\right) = 0$

$n + 4 = 0$

$n = - 4$

$n + 5 = 0$

$n = - 5$

$n = - 4 , - 15$