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We have two methods to find the values of c and nu2080-  1) Overestimate or simply take a large value for c and then find the value of  nu2080. 2) Underestimate or take a small value for nu2080 and then find the corresponding value of c. We must note that the choice of nu2080 and c is not unique. There can be many (actually, infinitely many) different combinations of nu2080 and c that would make the proof work. It depends on what inequalities you use while doing the upper/lower-bounding.  Now, the given argument is 2n^2 + 14n + 54 = O(n^2).  We are told to find the least integer c and then find the corresponding value of nu2080. The least integer possible will be 3 and c=3, now we find the corresponding value of nu2080. Inequality is 2n^2 + 14n + 54 <= c*n^2 2n^2 + 14n + 54 <= 3n^2 n^2 - 14n - 54 >= 0 Here b^2-4ac = 196- 4*1*(-54) = 196 + 216 = 412 which is greater than zero and hence roots are real.  Now, the roots of a quadratic equation can be given by -  x = [ -b +/- \sqrt{b^{2}-4 a c}] / 2a  x = [ 14 +/- \sqrt{412}] / 2 approximately, x = ( 14 +/- 20.2) / 2 = 34.2/2 , -6.2/2 = 17.1 , -3.1 Hence, the value of nu2080 must be greater than 17.1 or less than -3.1 Since, we need n>=nu2080, we take nu2080=18 and thus we get c=3 as the least possible integer.  We cannot use nu2080= -4 because we need n greater than nu2080 and not n less than nu2080.  Similarly we can solve the next part and find the value of nu2080 for c=24. The inequality becomes 2n^2 + 14n + 54 <= 24n^2 12n^2 - 14n - 54 >= 0 Here, b^2-4ac = 196 - 4*12*(-54) = 196 + 2592 = 2788 which is definitely grater than zero and hence the roots are real.  Now, the roots of a quadratic equation can be given by -  x = [ -b +/- \sqrt{b^{2}-4 a c}] / 2a  x = [ 14 +/- \sqrt{2788} ] / 2 Approximately, x = ( 14 +/- 52.8) / 2  x = 66.8 / 2 , -38.8 / 2 = 33.4, -19.4 Hence, the value of nu2080 must be greater than 33.4 or less than -19.4 Since we need n>=nu2080, we take the least possible value nu2080=34 for a given value of c=24. ...