## Elementary Algebra

Consecutive whole numbers have a difference of 1. Therefore, we let the first number be $x$ and the second number be $x+1$. The square of the first number is $x^{2}$ while the square of the second number is $(x+1)^{2}$. Since the sum of the squares of the two numbers is 145, we obtain: $x^{2}+(x+1)^{2}=145$ $x^{2}+x^{2}+1+2x=145$ $2x^{2}+2x+1-145=0$ $2x^{2}+2x-144=0$ $x^{2}+x-72=0$ $x^{2}-8x+9x-72=0$ $x(x-8)+9(x-8)=0$ $(x-8)(x+9)=0$ $x=8$ or $x=-9$ Disregarding the negative solution, the first unknown number is $8$. As a result, we obtain that the second unknown is; $8+1=9$.