1985 AHSME Problem 30

Let \lfloor x \rfloor be the greatest integer less than or equal to x. Then the number of real solutions to 4x^2-40\lfloor x \rfloor +51=0 is

\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 3 \qquad \mathrm{(E) \ }4

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.