Find the possible values of $p$ for which the equation has coincident roots.
Find the possible values of $p$ for which the equation $(2p+3)x^2+(4p-14)x+16p+1=0$ has coincident roots.
Coincident roots means 'equal roots'. For equal roots, we should use: $b^2-4ac=0$
$(4p-14)^2-4(2p+3)(16p+1)=0$
After solving this, I get: $-112p^2-536p+184=0$
I solved the above equation by this formula: $\frac{-b\pm \sqrt {b^2-4ac}}{2a}$
Is this right? Maybe there is some mistake but I can't find it. My answers are wrong. I got $-5.11$, $0.32$, my book says: $\frac{1}{2}$, $-3\frac{2}{7}$. Help. :(