# How do you solve #(x-3)^2-18=0#?

##### 2 Answers

#### Explanation:

The answer that was given above is absolutely correct. However, since the equation is given in the form

#(x - a)^2 + b = 0# ,

which is basically an intermediate step of the "completion of the circle" method, I would like to show you an easier way to solve it.

#color(white)(x)#

# (x - 3)^2 - 18 = 0#

... add

# <=> (x-3)^2 = 18#

Now, you can draw the root, but beware: there are two solutions, the negative and the positive one. (Please be sure that this is clear for you. As an example, for

# => x - 3 = sqrt(18) " or " x - 3 = -sqrt(18)#

... add

# => x = 3 + sqrt(18) " or " x = 3 - sqrt(18)#

So, your solution is

#x = 3 +- sqrt(18) = 3 +- sqrt(9 * 2) = 3 +- sqrt(9) * sqrt(2) = 3 +- 3 sqrt(2)#

Use the formula

So,

Now this is a Quadratic equation (in form

Use quadratic formula:

In this case