Question 816506
The sum of 128 and the square of its tens digit is 24 times its tens digit.
<pre>
Is that really what you meant?  With no mention of the units digit?

      128+tē = 24t
  tē-24t+128 = 0
 (t-8)(t-16) = 0

t-8 = 0;  t-16 = 0
  t = 8;     t = 16


So the tens digit is 8.  it could be any number with the tens digit 8.

It could be 80, 87, 286, 6784, or any number that has 8 for its next to
the last digit.  Are you sure you copied the problem right?

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Maybe you meant this:

The sum of 128 and the square of its tens digit is 24 times its UNITS digit.

Then 128+tē = 24u
         tē = 24u-128
          
The smallest t could be is 0 and the largest it could be is 9
So:
         0 &#8806; t &#8806; 9
so
        0 &#8806; tē &#8806; 81

       0 &#8806; 24u-128 &#8806; 81

     128 &#8806; 24u &#8806; 209

      5.3 < u < 8.7

So u can only be 6, 7 or 8   

If u = 6, then tē = 24u-128 = 24(6)-128=144-128 = 16, and t=4.
So one solution is 46.

If u = 7, then tē = 24u-128 = 24(7)-128=168-128 = 40, which has 
no integer square root.  So we discard u=7.

If u = 8, then tē = 24u-128 = 24(8)-128=192-128 = 64, and t=8.
So the only other solution is 88.

If the above was what you mean, the two solutions are 46 and 88.

Edwin</pre>