How to realize this transfer function with OP-AMPs?
Instead of looking at the problem as \$H(s)=\dfrac{1}{s + 0.383 + j*0.924}\$, take it as:
\$H(s)=\dfrac{s^2}{s^2 + 0.765*s + 1}\$  . Realizing this is easy, and was already done by Sallen and Key. Consider the following circuit and derivation of its transfer function:

By Kirchhoff's current law (KCL) applied at the \$Vx\$ node,
\$\dfrac{Vin-Vx}{Z_1}=\dfrac{Vx-Vout}{Z_3}+\dfrac{Vx-V^+}{Z_2}\$ (Eq.1),where \$V^+=V^-=Vout\$ because of the negative feedback.
Also, by applying KCL on OP-AMPs non-inverting input,
\$\dfrac{Vx-Vout}{Z_2}=\dfrac{Vout}{Z_4}\$ ,then: \$Vx=Vout*(\dfrac{Z_2}{Z_4}+1)\$
If we re-write all these in (Eq.1), then, we get:
\$\dfrac{Vin-Vout*(\dfrac{Z_2}{Z_4}+1)}{Z_1}=\dfrac{Vout*(\dfrac{Z_2}{Z_4}+1)-Vout}{Z_3}+\dfrac{Vout*(\dfrac{Z_2}{Z_4}+1)-Vout}{Z_2}\$
If we rearrange this equation, we will end up with:
\$\dfrac{Vout}{Vin}=\dfrac{Z_3*Z_4}{Z_1*Z_2+Z_3*(Z_1+Z_2)+Z_3*Z_4}\$
If \$Z_1\$ and \$Z_2\$ are selected as capacitors and others as resistors, we will have:
\$\dfrac{Vout}{Vin}=\dfrac{R_3*R_4}{\dfrac{1}{s^2*C_1*C_2}+\dfrac{R_3*(C_1+C_2)}{s*C_1*C_2}+R_3*R_4}\$
Rearranging this will give us a more meaningful equation:
\$\dfrac{Vout}{Vin}=\dfrac{s^2*R_3*R_4*C_1*C_2}{s^2*R_3*R_4*C_1*C_2+s*R_3*(C_1+C_2)+1}\$
Dividing numerator and denominator with \$R_3*R_4*C_1*C_2\$:
\$\dfrac{Vout}{Vin}=\dfrac{s^2}{s^2+s* \dfrac{R_3*(C_1+C_2)}{R_3*R_4*C_1*C_2}+\dfrac{1}{R_3*R_4*C_1*C_2}}\$
Simplifying this to:
\$\dfrac{Vout}{Vin}=\dfrac{s^2}{s^2+s*\dfrac{C_1+C_2}{R_4*C_1*C_2}+\dfrac{1}{R_3*R_4*C_1*C_2}}=\dfrac{s^2}{s^2 + 0.765*s + 1}\$
If we select \$C_1=C_2=1\text{F}\$, then \$R_4=2.6144\Omega\$ and \$R_3=0.3825\Omega\$.