Solve for b
b=2
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\left(\sqrt{6}\right)^{2}-\left(\sqrt{2}\right)^{2}=2b
Consider \left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\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}.
6-\left(\sqrt{2}\right)^{2}=2b
The square of \sqrt{6} is 6.
6-2=2b
The square of \sqrt{2} is 2.
4=2b
Subtract 2 from 6 to get 4.
2b=4
Swap sides so that all variable terms are on the left hand side.
b=\frac{4}{2}
Divide both sides by 2.
b=2
Divide 4 by 2 to get 2.
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