Solving quadratic equations without the quadratic forumla Is it possible to solve the following equation, in terms of $q$, without using the quadratic formula?
$t - (m-q)^2 = v - (m-p)^2$
I asked a similar question this morning (about the quadratic formula), but wondered if there was another approach, for the sake of completeness really!
Many thanks.
 A: Since our equation has very special form, we do not need the quadratic formula to "isolate" $q$.  
The equation has shape $A-(m-q)^2=B$. Rewrite it as $(m-q)^2=A-B$, and then as $q-m=\pm\sqrt{A-B}$, which yields $q=m\pm\sqrt{A-B}$.
A: Based off of what you had in your previous answer, here is a way to solve this without using the quadratic formula, although we will actually derive the quadratic formula in the process.
$$t-(m-q)^2 = v - (m-p)^2$$
Start out by moving everything except $(m-q)^2$ to the RHS and switching signs on both sides to obtain:
$$(m-q)^2 = (m-p)^2 + t - v$$
Now, $(m-q)^2 = (q-m)^2$, so we can rewrite this as
$$(q-m)^2 = (m-p)^2 + t - v$$
Then take the square root of both sides and we get
$$q-m = \pm \sqrt{(m-p)^2 + t - v)}$$
($\pm$ being shorthand for the two solutions, one with a $+$, one with a $-$)
And solving from here is a matter of moving $m$ to the other side.
But what you will see if you expand it out exactly as done in the previous question, is that in effect this derives the quadratic formula.
A: $(m-q)^2=t-v+(m-p)^2$, so:
 $m-q=\pm\sqrt{t-v+(m-p)^2}$ and finally: $q=m\mp\sqrt{t-v+(m-p)^2}$
