Question

How do you solve `x^2 + x - 12 = 0`?

Answer 1

The solutions are:

`x =3`

`x = -4`

Explanation:

`x^2 + x-12 =0`

The equation is of the form `color(blue)(ax^2+bx+c=0` where:

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

The Discriminant is given by:

`Delta=b^2-4*a*c`

`= (1)^2-(4*1*(-12))`

`= 1 +48 =49`

The solutions are found using the formula
`x=(-b+-sqrtDelta)/(2*a)`

`x = ((-1)+-sqrt(49))/(2*1) = (-1+-7)/2`

`x = (-1+7)/2 = 6/2 =3`

`x = (-1-7)/2 = -8/2 = -4`

The solutions are:

`x =3`

`x = -4`

Answer 2

`x=-4,3`

Explanation:

`x^2+x-12=0`

This is in the form of a Quadratic equation
(in form `ax^2+bx+c=0`)

Where

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

Use Quadratic formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Substitute the values

`rarrx=(-1+-sqrt(1^2-4(1)(-12)))/(2(1))`

`rarrx=(-1+-sqrt(1-(-48)))/(2)`

`rarrx=(-1+-sqrt(1+48))/(2)`

`rarrx=(-1+-sqrt49)/(2)`

`rarrx=(-1+-7)/2`

Now we have two values for `2`

`x=(-1-7)/2,x=(-1+7)/2`

If we solve for it,We get

`rArrx=-4,3`