Find two consecutive numbers whose squares have the sum 85.

Let two consecutive numbers be $x$ and $(x+1)$

Then according to question

$x^{2}+(x+1)^{2}=85$

$x^{2}+x^{2}+2 x+1=85$

$2 x^{2}+2 x-85+1=0$

$2 x^{2}+2 x-84=0$

$x^{2}+x-42=0$

$x^{2}+7 x-6 x-42=0$

$x(x+7)-6(x+7)=0$

$(x+7)(x-6)=0$

$(x+7)=0$

$x=-7$

Or

$(x-6)=0$

$x=6$

Since, x being a number,

Therefore,

When $x=-7$ then

$x+1=-7+1$

$=-6$

And when $x=6$ then

$x+1=6+1$

$=7$

Thus, two consecutive number be either 6,7 or $-6,-7$