Ok first divide both sides by 2 to get the radical by itself. You should end up with #x/2 - 3# on the left side and #sqrt(12-4x-x^2)# on the right. Then, you are going to square both sides, so you should get #((x/2)-3)^2# = #12-4x-x^2# . Now, we should know that #(a-b)^2# equals #(a^2 - 2ab + b^2)# , so the equation will be #((x^2)/4) - 3x +9# = #12-4x-x^2# . Let's combine the constants first, so let's subtract 9 from both sides. The new equation is #((x^2)/4) - 3x# = #3-4x-x^2#. Let's add 3x to both sides so we will get #(x^2)/4# = #3-x-x^2# . Let's multiply every term by 4 to get #x^2# = #12-4x-4x^2# and then add everything to one side to get #5x^2 + 4x -12# = 0. Now we can factor this quadratic into #(5x - 6) (x+2)# = 0. So either 5x - 6 = 0 or x+2=0. So our 2 possible answers are x=6/5 or x= -2. Remember to check for extraneous solutions.
