Did I solve the following quadratic equation correctly.
$$W(W+2)-7=2W(W-3)$$
I got.
$$W^2-8W+7$$ Then for my solution I got.
$$(W-1)(W-7)$$
Did I solve the following quadratic equation correctly.
$$W(W+2)-7=2W(W-3)$$
I got.
$$W^2-8W+7$$ Then for my solution I got.
$$(W-1)(W-7)$$
$$W(W+2)-7=2W(W-3)$$
$$\Downarrow$$
$$W^2+2W-7=2W^2-6W$$
$$\Downarrow$$
$$W^2-8W+7=0$$
$$\Downarrow$$
$$(W-1)(W-7)=0$$
You're right, well done. The solutions are $W=7$ and $W=1$.
EDIT: It should be written as I showed above, you've omitted "$=0$" part of the equation, intentionally or unintentionally.
$$W (W+2) -7 = 2W (W-3)$$ Expanding the brackets : $$W^2+2W -7 = 2W^2-6W$$ Then shift the right-hand side part of the equation to the left-hand side of the equation: $$W^2+2W-7-(2W^2-6W) = (2W^2-6W)-(2W^2-6W)$$ $$W^2+2W-7-(2W^2-6W)= 0$$ Expand the brackets: $$W^2+2W-7-2W^2+6W= 0$$ $$W^2-2W^2+2W+6W-7=0$$ $$-W^2+8W-7 =0$$ The above line is multiplied by (-1): $$(-1)(-W^2+8W-7) =(-1)(0)$$ Exapanding the brackets: $$W^2-8W+7 =0$$ This could be factorised as below $$W^2-7W-1W+7 =0$$ $$W(W-7)-1(W-7)=0$$ $$(W-1)(W-7) =0$$ Using Null-Factor law; Either $(W-1) =0$ or $(W-7) =0$
Therefore $W =1$ or $W =7$
So you have solved it correctly!