Evaluate
\frac{27-7\sqrt{5}}{22}\approx 0.515796553
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\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)}
Rationalize the denominator of \frac{7-\sqrt{5}}{7+\sqrt{5}} by multiplying numerator and denominator by 7-\sqrt{5}.
\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{7^{2}-\left(\sqrt{5}\right)^{2}}
Consider \left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{49-5}
Square 7. Square \sqrt{5}.
\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{44}
Subtract 5 from 49 to get 44.
\frac{\left(7-\sqrt{5}\right)^{2}}{44}
Multiply 7-\sqrt{5} and 7-\sqrt{5} to get \left(7-\sqrt{5}\right)^{2}.
\frac{49-14\sqrt{5}+\left(\sqrt{5}\right)^{2}}{44}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-\sqrt{5}\right)^{2}.
\frac{49-14\sqrt{5}+5}{44}
The square of \sqrt{5} is 5.
\frac{54-14\sqrt{5}}{44}
Add 49 and 5 to get 54.
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