Please help me solve for $L-L^2$?

Question

Unfortunately, I don't know any basic maths, and I need to solve the following equation for $L$ using the intercept of my graph:

$$\begin{aligned} intercept & = \frac{2L}{\pi^2 + L(L - 1)} \\ \end{aligned}$$

So far, I have:

$$\begin{aligned} intercept & = \frac{2L}{\pi^2 + L^2 - L} \\ \end{aligned}$$

$$\begin{aligned} intercept.\pi^2 + L^2 - L & = 2L \\ \end{aligned}$$

$$\begin{aligned} intercept.\pi^2 & = 2L - L^2 + L \\ \end{aligned}$$

$$\begin{aligned} intercept.\pi^2 & = 3L - L^2 \\ \end{aligned}$$

$$\begin{aligned} \frac{intercept.\pi^2}{3} & = L - L^2 \\ \end{aligned}$$

My excuse is that it's been a long time since I've done any basic maths, so I've probably rearranged it incorrectly... and I'm not sure what to do with the $L^2$, seeing as I need only one term for $L$ as I need to extract its value from the intercept of my graph. )

Answer 1

Let me call the intercept $n$. The main mistake you made is that when you clear the denominator on the right side, the entire denominator multiplies with the left side. So you get

$$n(\pi^2+L^2-L)=2L.$$

This can be rearranged into a quadratic equation:

$$n\pi^2+nL^2-nL=2L \\ nL^2+(-n-2)L+n\pi^2=0.$$

So the solutions, if there are any, can be found with the quadratic formula.

Answer 2

The thing is that $intercept$ multiplies the whole denominator. $$ intercept(\pi^2+L^2-L)=2L \\ intercept \times \pi^2+intercept \times L^2 -intercept \times L =2L \\ \pi^2=L-L^2+\frac{2L}{intercept} $$

Now keep in mind that the whole left hand side is a constant, say $\pi^2=c$.

Then, $$c=L-L^2+\frac{2L}{intercept}\\ -L^2+(1+\frac{2}{intercept})L-c=0$$

Now use the quadratic formula with $a=-1,b=(1+\frac{2}{intercept}),c=-\pi^2$ $$L_1,L_2=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

Now, you can get either 2, 1 or 0 roots, depending on the discriminant.