Question

How do you solve `3x^2 + 18x - 12 = 0`?

Answer 1

The solutions are
`color(blue)(x =(-9+sqrt(117))/3`

`color(blue)(x =(-9-sqrt(117))/3`

Explanation:

`3x^2+18x-12=0`

The equation is of the form `color(blue)(ax^2+bx+c=0` where:
`a=3, b=18, c=-12`

The Discriminant is given by:
`color(blue)(Delta=b^2-4*a*c`

`= (18)^2-(4*3*(-12))`

`= 324 +144=468`

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

`x = ((-18)+-sqrt(468))/(2*3) = (-18+-sqrt(468))/6`

(note:`sqrt468=sqrt(2*2*117)=2sqrt117`)

`x =(-18+-2sqrt(117))/6`

`x =(cancel2*(-9+-sqrt(117)))/cancel6`

`x =(-9+-sqrt(117))/3`

The solutions are
`color(blue)(x =(-9+sqrt(117))/3`

`color(blue)(x =(-9-sqrt(117))/3`

Answer 2

`x=-3+sqrt13`

`x=-3-sqrt13`

Explanation:

`3x^2+18x-12=0`

Divide both sides by `3`.

`x^2+6x-4=0` is a quadratic equation, `ax^2+bx+c`, where

`a=1, b=6, c=-4`

Solve by completing the square.

Add `4` to both sides of the equation.

`x^2+6x=4`

Divide `b` by `2` and square the result. Add to both sides of the equation.

`(6/2)^2=9`

`x^2+6x+9=4+9`

`x^2+6x+9=13`

Write the trinomial as a perfect square.

`(x+3)^2=13`

Take the square root of both sides.

`x+3=+-sqrt13`

Subtract `3` from both sides.

`x=-3+-sqrt13`

Solve for `x`.

`x=-3+sqrt13`

`x=-3-sqrt13`