Let the number be \( x \). According to the problem, we have the equation:
\[ x + x^2 = 90 \]
Rearranging this equation gives us:
\[ x^2 + x - 90 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 1 \), and \( c = -90 \). Plugging the values into the formula:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-90)}}{2 \cdot 1} \] \[ x = \frac{-1 \pm \sqrt{1 + 360}}{2} \] \[ x = \frac{-1 \pm \sqrt{361}}{2} \] \[ x = \frac{-1 \pm 19}{2} \]
This gives us two potential solutions:
\[ x = \frac{18}{2} = 9 \]
\[ x = \frac{-20}{2} = -10 \]
Thus, the two numbers that when added to their squares make 90 are \( 9 \) and \( -10 \).
To verify:
-
For \( x = 9 \): \[ 9 + 9^2 = 9 + 81 = 90 \]
-
For \( x = -10 \): \[ -10 + (-10)^2 = -10 + 100 = 90 \]
Both solutions are correct. Therefore, the numbers are \( 9 \) and \( -10 \).