# How do you solve the equation 3(x-2)^2-12=0

Feb 7, 2015

Lets simplify this scary thing 1st

This qill be $3 \left({x}^{2} + 4 - 4 x\right) - 12 = 3 {x}^{2} - 12 x$

so this is rewritten as$3 x \left(x - 4\right) = 0$

The zero rule here has to be either $x - 4 = 0 \mathmr{and} 3 x = 0$

Hence the roots of these equations are $x = 4 , 0$

Feb 8, 2015

Another path you could take to solving this equation would be to pull out a GCF. As you can see, all your terms on the left are divisible by 3. And since you just have 0 on the right, you can just divide everything by 3.

Then you get:

${\left(x - 2\right)}^{2} - 4 = 0$

Now you could expand the binomial here, but since 4 is a perfect square, it's just much easier to move it over to the other side (add 4 to both sides), and then take the square root of both sides of the equation.

You'd get:

$x - 2 = \pm 2$

Make sure you pay attention to the fact that since you took a square root, it's plus OR minus 2, and not just 2

Now, just add 2 to both sides, and you're done.

$x = 2 \pm 2 = 4 \mathmr{and} 0$

Hope that helped :)