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\frac{4\left(\sqrt{7}+3\right)}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}
Rationalize the denominator of \frac{4}{\sqrt{7}-3} by multiplying numerator and denominator by \sqrt{7}+3.
\frac{4\left(\sqrt{7}+3\right)}{\left(\sqrt{7}\right)^{2}-3^{2}}
Consider \left(\sqrt{7}-3\right)\left(\sqrt{7}+3\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\left(\sqrt{7}+3\right)}{7-9}
Square \sqrt{7}. Square 3.
\frac{4\left(\sqrt{7}+3\right)}{-2}
Subtract 9 from 7 to get -2.
-2\left(\sqrt{7}+3\right)
Divide 4\left(\sqrt{7}+3\right) by -2 to get -2\left(\sqrt{7}+3\right).
-2\sqrt{7}-6
Use the distributive property to multiply -2 by \sqrt{7}+3.