Is it possible to factor $(c+1)(c-\frac32)+1=0$ without expanding $(c+1)(c-\frac32)$?

Question

I have the equation $(c+1)(c-\frac32)+1=0$ and I'm trying to factor it without expanding $(c+1)(c-\frac32)$. I tried multiplying by $2$ to get rid of the fraction:

$$(c+1)(2c-3)+2=0$$ $$(c+1)(2c-3)+\frac25\times[2(c+1)-(2c-3)]=0$$ As you can see, I managed to make $2$ by using $c+1$ and $2c-3$. But I'm not sure if this helps.

Answer 1

If you prefer:

$$((c - 1/4) + 5/4)((c - 1/4) - 5/4)) + 1 = 0$$ $$\implies (c-1/4)^2 - 25/16 + 1 = 0$$ $$\implies (c-1/4)^2 = 9/16$$ $$\implies c-1/4 = 3/4, -3/4 \implies c = 1, -\frac 12$$

This is just completing the square, which is made easier by the two brackets in $(c+1)(c-3/2)$.

Answer 2

$$\begin{align}\begin{cases}c+1=-\frac{1}{a}\\c-\frac 32=a\end{cases}&\implies a+\frac1a=-\frac 52\\ &\implies a^2+\frac 52 a+1=0\\ &\implies a\in\left\{-\frac 12,-2\right\}\ \\ &\implies c\in\left\{1,-\frac 12\right\}.\end{align}$$