Question

How do you solve `2x^2+4x-30 =0`?

Answer 1

`x = -5, 3`

Explanation:

We have: `2 x^(2) + 4 x - 30 = 0`

Let's apply the quadratic formula:

`=> x = (- 4 pm sqrt(4^(2) - 4 (2) (- 30))) / (2 (2))`

`=> x = (- 4 pm sqrt(16 + 240)) / (4)`

`=> x = (- 4 pm sqrt(256)) / (4)`

`=> x = (- 4 pm 16) / (4)`

`=> x = - 1 pm 4`

`=> x = - 5, 3`

Therefore, the solutions to the equation are `x = - 5` and `x = 3`.

Answer 2

`x=-5 " or " x = 3`

Explanation:

All the terms in the equation are even, so we can divide by 2 immediately to make the numbers smaller.

`2x^2+ 4x -30 =0`

`x^2 +2x -15 = 0`

Try to factor before we use loner methods of the quadratic formula or completing the square.

Find factors of 15 which subtract (because of -15) to give 2.
The signs will be different (because of -15), there must be more positives. (because of +2)

5 and 3 will work! `5xx3 = 15 and 5-3 = 2`

`(x+5)(x-3) = 0`

Putting each factor equal to 0 gives the solutions:

`x=-5 " or " x = 3`