Let f(x)=x2+2ax+(3a+10). For α<1<β, since the leading coefficient is positive, we need f(1)<0.
f(1)=1+2a+3a+10=5a+11<0
a<−511
When f(1)<0 with positive leading coefficient, the parabola is negative at x=1, so 1 lies strictly between the two real roots. Discriminant is automatically positive.
a∈(−∞,−511).