find total resistance R1 and R2 which when connected in series have Req=40ohms and when in parellel Req=7.5ohms... please fast

Dear student,
When R1 and R2 are connected in series, the equivalent resistance is =R1+R2=40 ohm...(1)
R1=40-R2.....(2)
When R1 and R2 are connected in parallel, the equivalent resistance is
=   $$\begin{gathered} \frac{1}{Req}=\frac{1}{R1}+\frac{1}{R2} \\ \frac{1}{7.5}=\frac{1}{R1}+\frac{1}{R2} \end{gathered}$$

Putting the values of R1 from the equation(2), we get.

$$\begin{gathered} \frac{1}{7.5}=\frac{1}{40-R2}+\frac{1}{R2} \\ \frac{1}{7.5}=\frac{R2+40-R2}{R2(40-R2)} \\ \frac{1}{7.5}=\frac{40}{R2(40-R2)} \end{gathered}$$
40R2-R2^₂=300
R2^_2-40R2+300=0
R2^₂-10R2-30R2+300=0
R2(R2-10)-30(R2-10)=0
(R2-10)(R2-30)=0
R1=10ohm 
R2=30ohm.
Regards