+79 The question is to find the inverse of the following function - p(r) = 2r^2+2r-1
I realise that this cannot be factored by decomposition. So I'm using the quadratic formula. I've done most of the question but stuck near the end. I have noted the final answer on the side but i dont know how to get it to it. Can you help to finish this question? 
+9481 I followed your work..I think you've got it right so far!
(Except c = (-1 - y) , but you had it right in the problem.) 
You are very very close to the answer. Maybe this will nudge you in the right direction.
\(x=\frac{-2\,\pm\,2\sqrt{2y\,+\,3}}{4} \\~\\ x=\frac{2(-1\,\pm\,\sqrt{2y\,+\,3})}{2(2)}\) Factor a 2 out of the numerator and factor a 2 out of the denominator.
Below is the rest of the answer if you need more help.
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\(x=\frac{2(-1\,\pm\,\sqrt{2y\,+\,3})}{2(2)}\)
Divide the numerator and denominator by 2 .
\(x=\frac{-1\,\pm\,\sqrt{2y\,+\,3}}{2}\)
Rearrange the terms in the numerator.
\(x=\frac{\pm\,\sqrt{2y\,+\,3}\,-\,1}{2} \\~\\ \begin{array}\ x=\frac{\sqrt{2y\,+\,3}\,-\,1}{2} \qquad\text{ or }\qquad &x=\frac{-\,\sqrt{2y\,+\,3}\,-\,1}{2}& \\~\\ & x=\frac{-1(\,\sqrt{2y\,+\,3}\,+\,1)}{2}&\\~\\ & x=-\,\frac{\,\sqrt{2y\,+\,3}\,+\,1}{2}& \end{array}\)
Also...for your final answer, do you need to say p-1(r) for x and r for y ?
+130071 Thanks, hectictar.......here's another possible approach done by completing the square
Let's write "y" for p(r).......we'll come back to this later
y = 2r^2 + 2r -1 add 1 to both sides
y + 1 = 2r^2 + 2r
y + 1 = 2 [ r^2 + r] divide both sides by 2
[ y + 1] / 2 = r^2 + r completet the square on r
[ y + 1 ] / 2 + 1/4 + = r^2 + r + 1/4 simplify
[ 2y + 2 + 1 ] / 4 = ( r + 1/2)^2
[ 2y + 3 ] / 4 = ( r + 1/2)^2 take both roots
± √ [2y + 3] / 2 = r + 1/2
± √ [2y + 3] / 2 - 1/2 = r
± ( √ [2y + 3] - 1 ) / 2 = r "exchange" r and y
± ( √ [2r + 3] - 1 ) / 2 = y
For "y" write p-1(r)
p-1 (r) = ± ( √ [2r + 3] - 1 ) / 2
