Question

How do you solve by completing the square: `-2x^2-3x-3=0`?

Answer 1

`-2x^2-3x-3 = 0`

Divide all terms of both sides of the equation by `(-2)` so we are working with an expression that begins simply with `x^2`

`x^2 +3/2x + 3/2 = 0`

then move the constant `(-3/2)` off the left-side of the equation by subtracting `3/2` from both sides
`x^2+3/2x = -3/2`

To "complete the square" we need something of the form:
`(x+a)^2`
which equals `x^2+2ax +a^2`

Since we have computed the coefficient of `x` to be `3/2`
`rarr a = 3/4`
and `a^2 = 9/4`

Add `9/4` to both sides of the equation
`x^2+3/2x+9/4 = 9/4 - 3/2`

Rewrite the left-side as a square and simplify the right-side
`(x+3/2)^2 = 3/4`

Take the square root of both sides
`x+3/2 = +-sqrt(3)/2`

`x = -(3+sqrt(3))/2`
or
`x = -(3-sqrt(3))/2 = (sqrt(3)-3)/2`