Question

How do you find the roots for `x^2 + 2x – 48 = 0`?

Answer 1

Let us try completing the square

`0 = x^2+2x-48`

`= (x+1)^2 - 1 - 48 = (x+1)^2-49 = (x+1)^2 - 7^2`

So `(x+1)^2 = 7`

So `x+1=+-sqrt(7)`

So `x=-1+-sqrt(7)`

In general

`ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))`