Question

How do you solve `x^2-x=12`?

Answer 1

We can use the Sum-Product method.

Bring everything to one side:
`x^2-x=12->x^2-x-12=0`

We now have a quadratic equation of the form
`ax^2+bx+c=0` where `a=1, b=-1 and c=-12`

We find two numbers that will give `c=-12` as a product and `b=-1` as a sum (or difference).
We can try `1*12, 2*6,3*4`

`3and4` will fit, with a `-`sign to the `4`, as that would make the sum `3-4=-1` and the product `3*-4=-12`

Now we can rewrite the equation:
`(x+3)(x-4)=0`

Which leaves us with two possibilities:
`(x+3)=0->x=-3` OR `(x-4)=0->x=4`

Answer:
`x=-3 or x=4`