You can put this solution on YOUR website! 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