Question

The sum of the square of a positive number and the square of 2 more than the number is 130. what is the​ number

Answer 1

Let x be a positive number

$x^2+(x+2)^2=130 \ \ \ \ \ \ [x\ \textgreater \ 0] \\ x^2+x^2+4x+4-130=0 \\ 2x^2+4x-126=0 \\ x^2+2x-63=0 \\ D=b^2-4ac=2^2-4*1*(-63)=4+252=256 \\ x_{1,2}= \frac{-bб \sqrt{D} }{2a}= \frac{-2б \sqrt{256} }{2*1}= \frac{-2б16 }{2}\\x_1=-9 \ \ \ \ \O\\x_2=7$

Answer: the number is 7.

Check:
7² + (7+2)² = 49 + 9² = 49 + 81 = 130