is it possible to fit 6n(n-1) + 1 = 37 into the quadratic formula?

Question

This question results from my original question here.

I'm trying to find x for ... 6x(x-1) + 1 = 37 ... the answer is 3 but I need to know how to get to that answer. I've been advised to use the quadratic formula to do this and to use that I need something like:

ax2+bx+c=0

Here's my steps so far ...

6x(x-1) + 1 = 37
6x(x-1) + 1 - 37 = 0
6x(x-1) - 36 = 0

but now I'm stuck. Can I use the quadratic formula to solve this? What are the next steps to arrange this into the format ax2+bx+c=0? If this is not possible, how would I solve this? Many thanks

Answer 1

Note that $x(x-1)=x^{2}-x$ by the Distributive Property of Multiplication. Then, we have:

$6x(x-1)+1=37$

$6(x^{2}-x) - 36 = 0$

$x^{2}-x-6 = 0$

Can you find the solutions now? (Hint: Factoring works well.)

If not:

Because $x^{2}-x-6 = (x-3)(x+2)$, $x^{2}-x-6=0$ has the solutions $\boxed{x=-2,3.}$

Answer 2

$$6x(x-1) - 36 = 0 \Rightarrow 6x^2 - 6x - 36 = 0.$$ The quadratic formula states that $$x_i = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ are the solutions of $ax^2 + bx + c = 0.$ Therefore, $$x_i = \frac{-(-6) \pm \sqrt{36 - 4 \cdot 6 \cdot (-36)}}{2\cdot6}.$$ Alternatively, we can divide our original equation by $6$ to make our computations easier. That is, solving $x^2 - x - 6 = 0$ will give the same solutions as the original equation. Can you finish from here?