speed up is calculated by dividing the performance of Machine B by performance of machine A
We know the more time a process takes, the worse is it's performance. So, $\frac{P_B}{P_A}=\frac{T_A}{T_B}$
Where P stands for performance, and T stands for time taken.
Machine B runs floating-point instructions n times faster than machine A.
A spends 50 seconds in FP operations. 50 seconds in other operations.
B will spend $\frac{50}{n}$ seconds in FP operations, and 50 seconds in other operations (assuming equal time taken).
As per the question
$\frac{T_A}{T_B}=1.714$
=> $\frac{100}{(50/n)+50}=1.714$
After simplifying
$0.286n=1.714$
=> $n=5.99$
=> $n=6$