# The sum of the SQUARES of two consecutive positive integers is 145. How do you find the numbers?

Jan 27, 2016

n²+(n+1)²=145, =n²+n²+2n+1=145, 2n²+2n=144, n²+n=72, n²+n-72=0. n=(-b+-(b²-4*a*c))/2*a,( -1+(1-4*1*-72)^0.5)/2, =(-1+(289)^0.5)/2,=(-1+17)/2=8. n=8, n+1=9.

#### Explanation:

given.

Jan 27, 2016

I found $8 \mathmr{and} 9$

#### Explanation:

Let us call the numbers:

$n$
and
$n + 1$

we get (from our condition) that:

${\left(n\right)}^{2} + {\left(n + 1\right)}^{2} = 145$

rearrange and solve for $n$:
${n}^{2} + {n}^{2} + 2 n + 1 - 145 = 0$
$2 {n}^{2} + 2 n - 144 = 0$
${n}_{1 , 2} = \frac{- 2 \pm \sqrt{4 + 1152}}{4} = \frac{- 2 \pm 34}{4}$
${n}_{1} = - 9$
${n}_{2} = 8$
$n = 8$
$n + 1 = 9$