Here's an example of a 50 Ω source required to drive a 300 Ω load using two resistors. if you did the calculations, R1 would be 273.86 Ω and produces an attenuation into the 300 Ω load of 0.52277 V/V (5.63 dB). This is the benchmark to compare other impedance transformers with.

$$\boxed{R_1 = R_{L}\sqrt{1 - \dfrac{R_{IN}}{R_L}}}\hspace{0.3cm}\text{and}\hspace{0.3cm}\boxed{R_2 = \dfrac{R_{IN}\cdot R_L}{R_1}}$$

 RIN: Ω RL: Ω R1: Ω R2: Ω AV V/V

Circuit analysis

To find $$R_1$$ and $$R_2$$ we make the following simple observations about $$R_L$$ and $$R_{IN}$$: -

$$R_L = R_1 + \dfrac{R_2 R_{IN}}{R_2 + R_{IN}}\tag{1}$$

$$R_LR_2 + R_LR_{IN} = R_1R_2 + R_1R_{IN} + R_2R_{IN}$$

$$R_2 = \dfrac{R_{IN}(R_1 - R_L)}{R_L - R_1 - R_{IN}}\tag{3}$$

$$R_{IN} = \dfrac{R_2\cdot (R_1+R_L)}{R_1+R_2+R_L}\tag{2}$$

$$\text{Substituting (3) into (2) we get: -}$$

$$R_{IN} = \dfrac{\frac{R_{IN}(R_1-R_L)}{R_L-R_1-R_{IN}} \cdot(R_1+R_L)}{R_1+\frac{R_{IN}(R_1-R_L)}{R_L-R_1-R_{IN}}+R_L}\tag{4}$$

Then, by re-arranging equation (4) we get this: -

$$R_{IN}\hspace{1cm} = \hspace{1cm}\dfrac{R_{IN}\cdot (R_1^2 - R_L^2)}{(R_1+R_L)\cdot (R_L - R_1 - R_{IN})+R_1R_{IN} - R_LR_{IN}}$$

$$R_1R_L - R_1^2 - R_1R_{IN} + R_L^2 - R_1R_L - R_{IN}R_L + R_1R_{IN} - R_LR_{IN}\hspace{1cm} =\hspace{1cm} R_1^2 - R_L^2$$

$$-R_1^2 + R_L^2 - 2R_{IN}R_L = R_1^2 - R_L^2\hspace{1cm}\Longrightarrow\hspace{1cm} 2R_L^2 - 2R_1^2 = 2R_{IN}R_L$$

$$\boxed{\hspace{0.5cm}R_1 = R_L\sqrt{1 - \frac{R_{IN}}{R_L}}\hspace{0.5cm}}$$

To find $$R_2$$, solve equations (1) and (2) for $$R_1$$: -

$$R_1 = \dfrac{R_2R_L+R_LR_{IN}-R_2R_{IN}}{R_{IN}+R_2}\tag{1a}$$

$$R_1 = \dfrac{R_2R_L-R_LR_{IN}-R_2R_{IN}}{R_{IN}-R_2}\tag{2a}$$

Then equate them: -

$$\dfrac{R_2R_L+R_LR_{IN}-R_2R_{IN}}{R_{IN}+R_2}\hspace{1cm}=\hspace{1cm}\dfrac{R_2R_L-R_LR_{IN}-R_2R_{IN}}{R_{IN}-R_2}$$

$$(R_{IN}-R_2)\cdot(R_2R_L+R_LR_{IN}-R_2R_{IN})=(R_{IN}+R_2)\cdot(R_2R_L-R_LR_{IN}-R_2R_{IN})$$

$$R_2^2 R_L - R_L R_{IN}^2-R_2^2 R_{IN}\hspace{1cm}= \hspace{1cm}R_{IN}^2 R_L + R_2^2 R_{IN} - R_2^2 R_L$$

$$2R_2^2 R_L - 2R_2^2 R_{IN} = 2R_L R_{IN}^2\hspace{1cm}\Longrightarrow\hspace{1cm} R_2^2(R_L-R_{IN}) = R_L R_{IN}^2$$

$$\boxed{R_2\hspace{0.5cm} = \hspace{0.5cm}R_{IN}\sqrt{\dfrac{R_L}{R_L-R_{IN}}}\hspace{0.5cm} = \hspace{0.5cm}R_{IN}\dfrac{1}{\sqrt{1 - \dfrac{R_{IN}}{R_L}}}\hspace{0.5cm} = \hspace{0.5cm}\dfrac{R_{IN}\cdot R_L}{R_1}}$$