Question

How do you solve `3x ^ { 2} + 9x - 20= 0`?

Answer 1

`x = -3/2+-sqrt(321)/6`

Explanation:

We can use the quadratic formula or complete the square.

I happen to prefer completing the square, so that's the method presented here:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We will use this with `a = (6x+9)` and `b = sqrt(321)`

To cut down on arithmetic involving fractions, I will premultiply by `3*2^2 = 12`...

`0 = 12(3x^2+9x-20)`

`color(white)(0) = 36x^2+108x-240`

`color(white)(0) = (6x)^2+2(6x)(9)+(9)^2-321`

`color(white)(0) = (6x+9)^2-(sqrt(321))^2`

`color(white)(0) = ((6x+9)-sqrt(321))((6x+9)+sqrt(321))`

`color(white)(0) = (6x+9-sqrt(321))(6x+9+sqrt(321))`

Hence:

`x = 1/6(-9+-sqrt(321)) = -3/2+-sqrt(321)/6`