Let the unknown number be \( x \). According to the problem, the square of the number is 22 less than 13 times the number. We can set up the equation as follows:
\[ x^2 = 13x - 22 \]
Now, rearranging the equation gives us:
\[ x^2 - 13x + 22 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = -13 \), and \( c = 22 \).
First, we calculate the discriminant:
\[ b^2 - 4ac = (-13)^2 - 4 \cdot 1 \cdot 22 = 169 - 88 = 81 \]
Since the discriminant is positive, we will have two real solutions. Now we substitute back into the quadratic formula:
\[ x = \frac{-(-13) \pm \sqrt{81}}{2 \cdot 1} = \frac{13 \pm 9}{2} \]
This gives us two potential solutions:
- \( x = \frac{13 + 9}{2} = \frac{22}{2} = 11 \)
- \( x = \frac{13 - 9}{2} = \frac{4}{2} = 2 \)
Thus, the possible values of \( x \) are \( 11 \) and \( 2 \).
To confirm, we can check both values:
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For \( x = 11 \): \[ x^2 = 11^2 = 121 \] \[ 13x - 22 = 13 \cdot 11 - 22 = 143 - 22 = 121 \] This is correct.
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For \( x = 2 \): \[ x^2 = 2^2 = 4 \] \[ 13x - 22 = 13 \cdot 2 - 22 = 26 - 22 = 4 \] This is also correct.
Thus, the numbers that satisfy the condition are \( \boxed{11} \) and \( \boxed{2} \).
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