Solve {left(1+2sqrt{3}right)}^2-4(sqrt{3}-3)

$${ \left(1+2 \sqrt{ 3 } \right) }^{ 2 } -4( \sqrt{ 3 } -3)$$

Evaluate

$$25$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+2\sqrt{3}\right)^{2}.

$$1+4\sqrt{3}+4\left(\sqrt{3}\right)^{2}-4\left(\sqrt{3}-3\right)$$

The square of \sqrt{3} is 3.

$$1+4\sqrt{3}+4\times 3-4\left(\sqrt{3}-3\right)$$

Multiply 4 and 3 to get 12.

$$1+4\sqrt{3}+12-4\left(\sqrt{3}-3\right)$$

Add 1 and 12 to get 13.

$$13+4\sqrt{3}-4\left(\sqrt{3}-3\right)$$

Use the distributive property to multiply -4 by \sqrt{3}-3.

$$13+4\sqrt{3}-4\sqrt{3}+12$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$13+12$$

Add 13 and 12 to get 25.

$$25$$

Factor

$$5^{2}$$