Question

Solve using any method -x^2+x+12=0

Answer 1

For this case we have the following quadratic equation:

`-x ^ 2 + x + 12 = 0`

We can solve it by:

`x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}`

Where:

`a = -1\\b = 1\\c = 12`

Substituting:

`x = \frac {-1 \pm \sqrt {1 ^ 2-4 (-1) (12)}} {2 (-1)}\\x = \frac {-1 \pm \sqrt {1 + 48}} {- 2}\\x = \frac {-1 \pm \sqrt {49}} {- 2}\\x = \frac {-1 \pm \sqrt {49}} {- 2}\\x = \frac {-1 \pm7} {- 2}`

We have two roots:

`x_ {1} = \frac {-1 + 7} {- 2} = \frac {6} {- 2} = - 3\\x_ {2} = \frac {-1-7} {- 2} = \frac {-8} {- 2} = 4`

Answer:

`x_ {1} = - 3\\x_ {2} = 4`

Answer 2

`x_1=4\\x_2=-3`

Explanation:

Multiply both sides of the equation by -1:

`(-1)(-x^2+x+12)=0(-1)\\x^2-x-12=0`

Now, you can Factor the quadratic equation to solve it:

Choose two number that when you multiply them you get -12 and when you add them you get -1. These numbers are: -4 and 3

Therefore, you get:

`(x-4)(x+3)=0\\x_1=4\\x_2=-3`