Let the two consecutive numbers be $x$ and $x+1$.

The sum of the squares of these numbers is given by: $x^2 + (x+1)^2 = 145$

First, expand the equation: $x^2 + (x^2 + 2x + 1) = 145$

Combine like terms: $2x^2 + 2x + 1 = 145$

Subtract 145 from both sides to set the equation to zero: $2x^2 + 2x + 1 - 145 = 0$ $2x^2 + 2x - 144 = 0$

Divide the entire equation by 2 to simplify: $x^2 + x - 72 = 0$

Next, solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 1$, and $c = -72$.

Calculate the discriminant: $b^2 - 4ac = 1^2 - 4(1)(-72) = 1 + 288 = 289$

Find the square root of the discriminant: $\sqrt{289} = 17$

Apply the quadratic formula: $x = \frac{-1 \pm 17}{2}$

This gives us two solutions: $x = \frac{-1 + 17}{2} = \frac{16}{2} = 8$ $x = \frac{-1 - 17}{2} = \frac{-18}{2} = -9$

Therefore, the two pairs of consecutive numbers are:

1. $x = 8$ and $x+1 = 9$
2. $x = -9$ and $x+1 = -8$

So, the numbers are either $(8, 9)$ or ( (-9,