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

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

Evaluate

$$-\sqrt{3}-\frac{1}{4}\approx -1.982050808$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

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

Since -\frac{\sqrt{3}}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.

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

To raise \frac{-\sqrt{3}+2}{2} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2^{2}}{2^{2}}.

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

Since \frac{\left(-\sqrt{3}+2\right)^{2}}{2^{2}} and \frac{2\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

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

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

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

The square of \sqrt{3} is 3.

$$\frac{3-4\sqrt{3}+4}{2^{2}}-2$$

Add 3 and 4 to get 7.

$$\frac{7-4\sqrt{3}}{2^{2}}-2$$

Calculate 2 to the power of 2 and get 4.

$$\frac{7-4\sqrt{3}}{4}-2$$

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

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

Since \frac{7-4\sqrt{3}}{4} and \frac{2\times 4}{4} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 7-4\sqrt{3}-2\times 4.

$$\frac{7-4\sqrt{3}-8}{4}$$

Do the calculations in 7-4\sqrt{3}-8.

$$\frac{-1-4\sqrt{3}}{4}$$

Expand

$$-\sqrt{3}-\frac{1}{4}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

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

Since -\frac{\sqrt{3}}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.

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

To raise \frac{-\sqrt{3}+2}{2} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2^{2}}{2^{2}}.

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

Since \frac{\left(-\sqrt{3}+2\right)^{2}}{2^{2}} and \frac{2\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

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

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

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

The square of \sqrt{3} is 3.

$$\frac{3-4\sqrt{3}+4}{2^{2}}-2$$

Add 3 and 4 to get 7.

$$\frac{7-4\sqrt{3}}{2^{2}}-2$$

Calculate 2 to the power of 2 and get 4.

$$\frac{7-4\sqrt{3}}{4}-2$$

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

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

Since \frac{7-4\sqrt{3}}{4} and \frac{2\times 4}{4} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 7-4\sqrt{3}-2\times 4.

$$\frac{7-4\sqrt{3}-8}{4}$$

Do the calculations in 7-4\sqrt{3}-8.

$$\frac{-1-4\sqrt{3}}{4}$$