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\frac{\left(2\sqrt{3}-4\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}
Rationalize the denominator of \frac{2\sqrt{3}-4\sqrt{2}}{\sqrt{3}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}+\sqrt{2}.
\frac{\left(2\sqrt{3}-4\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}
Consider \left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\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(2\sqrt{3}-4\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{3-2}
Square \sqrt{3}. Square \sqrt{2}.
\frac{\left(2\sqrt{3}-4\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{1}
Subtract 2 from 3 to get 1.
\left(2\sqrt{3}-4\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)
Anything divided by one gives itself.
2\left(\sqrt{3}\right)^{2}+2\sqrt{3}\sqrt{2}-4\sqrt{2}\sqrt{3}-4\left(\sqrt{2}\right)^{2}
Apply the distributive property by multiplying each term of 2\sqrt{3}-4\sqrt{2} by each term of \sqrt{3}+\sqrt{2}.
2\times 3+2\sqrt{3}\sqrt{2}-4\sqrt{2}\sqrt{3}-4\left(\sqrt{2}\right)^{2}
The square of \sqrt{3} is 3.
6+2\sqrt{3}\sqrt{2}-4\sqrt{2}\sqrt{3}-4\left(\sqrt{2}\right)^{2}
Multiply 2 and 3 to get 6.
6+2\sqrt{6}-4\sqrt{2}\sqrt{3}-4\left(\sqrt{2}\right)^{2}
To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.
6+2\sqrt{6}-4\sqrt{6}-4\left(\sqrt{2}\right)^{2}
To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.
6-2\sqrt{6}-4\left(\sqrt{2}\right)^{2}
Combine 2\sqrt{6} and -4\sqrt{6} to get -2\sqrt{6}.
6-2\sqrt{6}-4\times 2
The square of \sqrt{2} is 2.
6-2\sqrt{6}-8
Multiply -4 and 2 to get -8.
-2-2\sqrt{6}
Subtract 8 from 6 to get -2.