Solve (sqrt{3}-2)^2+3/2sqrt{3}1/8

$$( \sqrt { 3 } - 2 ) ^ { 2 } + \frac { 3 } { 2 \sqrt { 3 } } \frac { 1 } { 8 }$$

Evaluate

$$-\frac{63\sqrt{3}}{16}+7\approx 0.180049945$$

Short Solution Steps

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

$$\left(\sqrt{3}\right)^{2}-4\sqrt{3}+4+\frac{3}{2\sqrt{3}}\times \frac{1}{8}$$

The square of \sqrt{3} is 3.

$$3-4\sqrt{3}+4+\frac{3}{2\sqrt{3}}\times \frac{1}{8}$$

Add 3 and 4 to get 7.

$$7-4\sqrt{3}+\frac{3}{2\sqrt{3}}\times \frac{1}{8}$$

Rationalize the denominator of \frac{3}{2\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$7-4\sqrt{3}+\frac{3\sqrt{3}}{2\left(\sqrt{3}\right)^{2}}\times \frac{1}{8}$$

The square of \sqrt{3} is 3.

$$7-4\sqrt{3}+\frac{3\sqrt{3}}{2\times 3}\times \frac{1}{8}$$

Cancel out 3 in both numerator and denominator.

$$7-4\sqrt{3}+\frac{\sqrt{3}}{2}\times \frac{1}{8}$$

Multiply \frac{\sqrt{3}}{2} times \frac{1}{8} by multiplying numerator times numerator and denominator times denominator.

$$7-4\sqrt{3}+\frac{\sqrt{3}}{2\times 8}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 7-4\sqrt{3} times \frac{2\times 8}{2\times 8}.

$$\frac{\left(7-4\sqrt{3}\right)\times 2\times 8}{2\times 8}+\frac{\sqrt{3}}{2\times 8}$$

Since \frac{\left(7-4\sqrt{3}\right)\times 2\times 8}{2\times 8} and \frac{\sqrt{3}}{2\times 8} have the same denominator, add them by adding their numerators.

$$\frac{\left(7-4\sqrt{3}\right)\times 2\times 8+\sqrt{3}}{2\times 8}$$

Do the multiplications in \left(7-4\sqrt{3}\right)\times 2\times 8+\sqrt{3}.

$$\frac{112-64\sqrt{3}+\sqrt{3}}{2\times 8}$$

Do the calculations in 112-64\sqrt{3}+\sqrt{3}.

$$\frac{112-63\sqrt{3}}{2\times 8}$$

Expand 2\times 8.

$$\frac{112-63\sqrt{3}}{16}$$

Factor

$$\frac{7 {(16 - 9 \sqrt{3})}}{16} = 0.18004994519754625$$