# How do you solve this quadratic equation for completing the square?: 3x^2 -12x = -4

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Hi, Rosey,

For this one, if you follow the form of ax^2 + bx + c = 0, we already have c on the other side. And, the method I like to use, "a" has to be 1. So, the first thing we would do is divide each side by 3. That gives us:

x^2 - 4x = -4/3

Then, we take half of 4 (with the 4x) and square that result. Half of 4 is 2. Squaring that is 4. So, we now add 4 to each side, giving us:

x^2 - 4x + 4 = 8/3

That allows the left side to factor into (x-2)^2

(x-2)^2 = 8/3

taking the square root of each side

x-2 = +-sqrt(8/3)

Adding 2 to each side:

x = 2 +- sqrt(8/3)

From here, a variety of things could go on. If there is anything, we simplify the square root:

sqrt(8/3) = sqrt8 / sqrt3 = sqrt4 * sqrt2 / sqrt3

= 2 * sqrt2 / sqrt3 * sqrt3/sqrt3 = 2 sqrt6 /3

So, then, we have:

2 +- 2sqrt6 /3

We can go further with this, making it:

(6 +- 2sqrt6)/3

Good luck, Rosey. I hope this is understandable.

Till Then,

Steve

By completing the square, solve: `3x^2-12x = -4.`

First, factor a 3 from left side of equation to give:

`3(x^2 - 4x) = -4` In order to complete square we need to find value of "c" in `ax^2 + bx + c` .

`3(x^2 - 4x + c) = -4 + 3c` We need to multiply by 3 on left because we would need to distribute 3 from left.

In orderto find "c", we divide b by 2 and then square it.

`3(x^2 - 4x + 4) = -4 + 12` Factor next

`3(x - 2)^2 = 8` Divide both sides by 3.

`(x - 2)^2 = 8/3` Take square root of both sides.

`x - 2 =+-sqrt(8/3)` Add 2 to both sides.

`x = 2+-sqrt(8/3)` May need to simplify `sqrt(8/3) = (2sqrt(6))/3`

**Final answer:** `x = 2+-(2sqrt(6))/3`

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