Find two consecutive integers such that their product is 11 more than 2 times their sum

I strongly recommend starting out by choosing an appropriate variable.
We could let the first integer be x and the next consecutive integer be x+1.

The sum of these two consec. integers is x + (x+1), or 2x+1.  Twice this sum is 4x+2.

The product of these two consec. int. is x(x+1), which is 11 more than twice their sum.  Writing this out symbolically,

x(x+1) = 2[2x+1] + 11

Expanding:  x^2 + x = 11 + 4x + 2
                    x^2 + x - 4x - 13 = 0
                     x^2 - 3x - 13 = 0

Unfortunately, this quadratic equation, while solvable, has neither integer nor real roots.  Thus, there is no solution which consists of the integer x and the next consecute integer x+1.  Are you positive you have copied down this problem correctly?

I’m guessing that the answer is c, but i’m not sure which grid is being talked about. since 5x5= 25 and 3x3= 9 (which there are no answers for). a is not right because while the ratio is the same, 6/25= 0.24 which is equal to 24% not 30%.