Question

How do you solve by completing the square: `ax^2+bx+c=0`?

Answer 1

`ax^2 + bx +c = 0`

Divide all terms by `a` so as to reduce the coefficient of `x^2` to `1`
`x^2 +b/a x + c/a=0`

Subtract the constant term from both sides of the equation
`x^2+b/a x = - c/a`

To have a square on the left side the third term (constant) should be
`(b/(2a))^2`

So add that amount to both sides
`x^2 +b/a x +(b/(2a))^2 = (b/(2a))^2 - c/a`

Re-write the left-side as a square
`(x+(b/(2a)))^2 = (b/(2a))^2 -c/a`

Take the square root of both sides (remembering that the result could be plus or minus)
`x+(b/(2a)) = +-sqrt((b/(2a))^2 -c/a)`

Subtract the constant term on the left side from both sides
`x = +-sqrt((b/(2a))^2 -c/a) - (b/(2a))`

or, with some simplification

`x= (-b+-sqrt(b^2-4ac))/(2a)`

(the standard form for solving a quadratic)