Curve $C$ has an equation of : $$y = 1 + \frac {2x+p}{(x-2)(x+3)} $$ where p is a constant
Find the range of values of p for which $C$ has more than one stationary point
I answered $$y'= \frac{2(x^2+x-6)-(2x+p)(2x+1)}{(x^2+x-6)^2}$$ $$y'= 0$$ so $$0 = 2(x^2+x-6)-(2x+p)(2x+1)$$ $$0 = 2x^2+2px+p+12$$
what to do next?
Then you have to solve $$2x^2+2px+p+12=0$$ use the quadratic formula for $$ax^2+bx+c=0$$ this is $$x_{1,2}=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$$