Seventy-five times an integer, minus 36, equals 21 times the square of the integer. Which equation could be used to solve for the unknown integer?

75n-36=21n^2 rearrange so we have an equation equal to zero ...

21n^2-75n+36=0 divide whole equation by 3

7n^2-25n+12=0 now factor ...

7n^2-21n-4n+12=0

7n (n-3) - 4 (n-3) = 0

(7n-4) (n-3) = 0

Since this is an integer 4/7 is extraneous.

The integer value is 3