There is an algebraic way to solve this question. You can solve it the way you were doing but with one change.
you did really well up until \(\displaystyle w^2-6w=7\)
There is a formula that is called the quadratic formula that you can use to solve for w. you must use this equation because the w is squared and you have and extra w. The general form to this equation is
\(\displaystyle w=\frac{-b_-^+\sqrt{b^2-4ac}}{2a}\)
But you need to get you equation in the general \(\displaystyle aw^2+bw+c=0\)
Where a,b, and c are numbers in front of your variables.
when you put your equation in the proper form it reads \(\displaystyle w^2-6w-7=0\)
so your a=1....b=-6....c=-7
when we plug it in to the quadratic formula you get
\(\displaystyle w=\frac{-(-6)_-^+\sqrt{(-6)^2-(4*1*-7)}}{2(1)}\)
Now all you have to do is the basic math rules to solve for w and because you have a squared term in your equation you will have two answers.
You can see this in you quadratic formula with its \(\displaystyle _-^+\)... which is a plus or minus.
this gives w=7 and W=-1
but there is another way to do it using some other algebraic rules that will make it go quicker, check it out.
Your original equation was\(\displaystyle (w-3)^2=16\)
1. square root both both sides which gives
\(\displaystyle w-3=\sqrt{16}=4\)
2. add three to both sides giving
\(\displaystyle w=4+3=7\)
You get your answer. You will get w=-1 if you follow the square root rules and put a \(\displaystyle _-^+\) infront of the \(\displaystyle \sqrt{16}\)
Is that clear for you?
