Biased coin maximising one run of exactly two heads

You may choose the heads probability $p$ of a coin, which is then tossed $4$ times. You want to maximise the probability that the sequence contains a block of exactly two heads in a row, with no run of three or more heads. So $HHTH$ and $HHTT$ qualify, while $HHHT$ and $HHHH$ do not. The optimal $p$ has the form $\dfrac{a-\sqrt b}{c}$ with integers $a,b,c$ and $c$ as small as possible. Find $a+b+c$.

$8$. The qualifying patterns are $HHT\ast$, $THHT$ and $\ast THH$, which together have probability $p^2(1-p)(3-p)=3p^2-4p^3+p^4$. Its derivative $2p(2p^2-6p+3)$ vanishes inside $(0,1)$ at $p=\tfrac{3-\sqrt3}{2}\approx0.634$, so $a+b+c=3+3+2=8$.