# How do you find two consecutive even integers such that the square of the smaller is 10 more than the larger?

Jun 21, 2016

I found: $4 \mathmr{and} 6$

#### Explanation:

Call the two numbers:
$2 n$
and
$2 n + 2$
so we get:
${\left(2 n\right)}^{2} = \left(2 n + 2\right) + 10$
$4 {n}^{2} = 2 n + 2 + 10$
$4 {n}^{2} - 2 n - 12 = 0$
solve for $n$:
${n}_{1 , 2} = \frac{2 \pm \sqrt{4 + 192}}{8} =$
${n}_{1 , 2} = \frac{2 \pm 14}{8}$
${n}_{1} = 2$
${n}_{2} = - \frac{12}{8} = - \frac{3}{2}$
so we have using ${n}_{1}$:

$4 \mathmr{and} 6$