<PRE><B>Question 253742</B>
Let the required number be x.
Given, four times the square of a number is 21 more than eight times the number
    4x^2= 8x+21
    4x^2-8x-21 =0
    4x^2-14x+6x-21 = 0
    2x(2x-7)+3(2x-7) = 0
    (2x-7)(2x+3) = 0
           2x-7 =0   or  2x+3 = 0
             2x = 7  or  2x = -3 
              x = 7/2 or  x = -3/2</PRE>