Algebra - Factoring Quadratic Equations

I'm in algebra and this problem was under the lesson Factoring to Solve Quadratic Equations. The problem is the following:

The product of two consecutive numbers is 14 less than 10 times the smaller number. Find each number.

I've come up with $$x$$ and $$10x-14$$. I know that there should be two variables, but that's all I could come up with as far as an expression goes. If anyone can give me a better expression or teach me how to solve this it'd be greatly appreciated. Thank you.

2 Answers

Consecutive integers are x and x+1.

Product is $$x(x+1)$$

Thus the equation to solve is $$x(x+1)=10x-14$$

Call the smaller number $$x$$. Then we have

$$x(x+1) +14 = 10x$$

$$x^2 -9x + 14 = 0$$

$$(x -7)(x-2) = 0$$

so the solutions are $$x=7$$ or $$x=2$$ and so the smaller number is either $$2$$ or $$7$$. The larger number will obviously be $$3$$ or $$8$$ respectively.