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\frac{4\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right)}{\left(2\sqrt{3}+\sqrt{14}\right)\left(2\sqrt{3}-\sqrt{14}\right)}
Rationalize the denominator of \frac{4\sqrt{5}}{2\sqrt{3}+\sqrt{14}} by multiplying numerator and denominator by 2\sqrt{3}-\sqrt{14}.
\frac{4\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right)}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{14}\right)^{2}}
Consider \left(2\sqrt{3}+\sqrt{14}\right)\left(2\sqrt{3}-\sqrt{14}\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{4\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right)}{2^{2}\left(\sqrt{3}\right)^{2}-\left(\sqrt{14}\right)^{2}}
Expand \left(2\sqrt{3}\right)^{2}.
\frac{4\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right)}{4\left(\sqrt{3}\right)^{2}-\left(\sqrt{14}\right)^{2}}
Calculate 2 to the power of 2 and get 4.
\frac{4\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right)}{4\times 3-\left(\sqrt{14}\right)^{2}}
The square of \sqrt{3} is 3.
\frac{4\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right)}{12-\left(\sqrt{14}\right)^{2}}
Multiply 4 and 3 to get 12.
\frac{4\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right)}{12-14}
The square of \sqrt{14} is 14.
\frac{4\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right)}{-2}
Subtract 14 from 12 to get -2.
-2\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right)
Divide 4\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right) by -2 to get -2\sqrt{5}\left(2\sqrt{3}-\sqrt{14}\right).
-4\sqrt{3}\sqrt{5}+2\sqrt{5}\sqrt{14}
Use the distributive property to multiply -2\sqrt{5} by 2\sqrt{3}-\sqrt{14}.
-4\sqrt{15}+2\sqrt{5}\sqrt{14}
To multiply \sqrt{3} and \sqrt{5}, multiply the numbers under the square root.
-4\sqrt{15}+2\sqrt{70}
To multiply \sqrt{5} and \sqrt{14}, multiply the numbers under the square root.