Question

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

Answer 1

Answer:

$x_1=4\\x_2=-3$

Step-by-step 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:

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

Answer 2

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:

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