Differentiability at $x=1$ requires both continuity and matching derivatives.

Continuity at $x=1$:

$2\alpha (1-2)+2\beta =(\alpha +3)+(\alpha -\beta )$

$-2\alpha +2\beta =2\alpha -\beta +3\Rightarrow 4\alpha -3\beta =-3\dots (1)$

Derivative match at $x=1$: $f_{-}^{\prime }(x)=4\alpha x+2\beta$, $f_{+}^{\prime }(x)=\alpha +3$.

$4\alpha +2\beta =\alpha +3\Rightarrow 3\alpha +2\beta =3\dots (2)$

Solving (1) and (2): multiply (1) by 2, (2) by 3, then add: $17\alpha =3\Rightarrow \alpha =\frac{3}{17}$.

From (2): $\beta =\frac{3-9/17}{2}=\frac{21}{17}$.

$\alpha +\beta =\frac{24}{17}\Rightarrow 34(\alpha +\beta )=48$.