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

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

Evaluate

$$28-16\sqrt{3}\approx 0.287187079$$

Solution Steps

Calculate 4 to the power of 2 and get 16.

$$16+\left(2\sqrt{3}\right)^{2}-2\times 2\times 4\sqrt{3}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$16+2^{2}\left(\sqrt{3}\right)^{2}-2\times 2\times 4\sqrt{3}$$

Calculate 2 to the power of 2 and get 4.

$$16+4\left(\sqrt{3}\right)^{2}-2\times 2\times 4\sqrt{3}$$

The square of \sqrt{3} is 3.

$$16+4\times 3-2\times 2\times 4\sqrt{3}$$

Multiply 4 and 3 to get 12.

$$16+12-2\times 2\times 4\sqrt{3}$$

Add 16 and 12 to get 28.

$$28-2\times 2\times 4\sqrt{3}$$

Multiply 2 and 2 to get 4.

$$28-4\times 4\sqrt{3}$$

Multiply 4 and 4 to get 16.

$$28-16\sqrt{3}$$

Factor

$$4 {(7 - 4 \sqrt{3})} = 0.287187079$$