Solve (frac{-sqrt{21}-1}{2})^2-6

$$( \frac { - \sqrt { 21 } - 1 } { 2 } ) ^ { 2 } - 6$$

Evaluate

$$\frac{\sqrt{21}-1}{2}\approx 1.791287847$$

Short Solution Steps

To raise \frac{-\sqrt{21}-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-\sqrt{21}-1\right)^{2}}{2^{2}}-6$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 6 times \frac{2^{2}}{2^{2}}.

$$\frac{\left(-\sqrt{21}-1\right)^{2}}{2^{2}}-\frac{6\times 2^{2}}{2^{2}}$$

Since \frac{\left(-\sqrt{21}-1\right)^{2}}{2^{2}} and \frac{6\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(-\sqrt{21}-1\right)^{2}-6\times 2^{2}}{2^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-\sqrt{21}-1\right)^{2}.

$$\frac{\left(-\sqrt{21}\right)^{2}-2\left(-\sqrt{21}\right)+1}{2^{2}}-6$$

Calculate -\sqrt{21} to the power of 2 and get \left(\sqrt{21}\right)^{2}.

$$\frac{\left(\sqrt{21}\right)^{2}-2\left(-\sqrt{21}\right)+1}{2^{2}}-6$$

Multiply -2 and -1 to get 2.

$$\frac{\left(\sqrt{21}\right)^{2}+2\sqrt{21}+1}{2^{2}}-6$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{21}\right)^{2}+2\sqrt{21}+1}{4}-6$$

The square of \sqrt{21} is 21.

$$\frac{21+2\sqrt{21}+1}{4}-6$$

Add 21 and 1 to get 22.

$$\frac{22+2\sqrt{21}}{4}-6$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 6 times \frac{4}{4}.

$$\frac{22+2\sqrt{21}}{4}-\frac{6\times 4}{4}$$

Since \frac{22+2\sqrt{21}}{4} and \frac{6\times 4}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{22+2\sqrt{21}-6\times 4}{4}$$

Do the multiplications in 22+2\sqrt{21}-6\times 4.

$$\frac{22+2\sqrt{21}-24}{4}$$

Do the calculations in 22+2\sqrt{21}-24.

$$\frac{-2+2\sqrt{21}}{4}$$

Expand

$$\frac{\sqrt{21} - 1}{2} = 1.79128784747792$$

Short Solution Steps

To raise \frac{-\sqrt{21}-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-\sqrt{21}-1\right)^{2}}{2^{2}}-6$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 6 times \frac{2^{2}}{2^{2}}.

$$\frac{\left(-\sqrt{21}-1\right)^{2}}{2^{2}}-\frac{6\times 2^{2}}{2^{2}}$$

Since \frac{\left(-\sqrt{21}-1\right)^{2}}{2^{2}} and \frac{6\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(-\sqrt{21}-1\right)^{2}-6\times 2^{2}}{2^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-\sqrt{21}-1\right)^{2}.

$$\frac{\left(-\sqrt{21}\right)^{2}-2\left(-\sqrt{21}\right)+1}{2^{2}}-6$$

Calculate -\sqrt{21} to the power of 2 and get \left(\sqrt{21}\right)^{2}.

$$\frac{\left(\sqrt{21}\right)^{2}-2\left(-\sqrt{21}\right)+1}{2^{2}}-6$$

Multiply -2 and -1 to get 2.

$$\frac{\left(\sqrt{21}\right)^{2}+2\sqrt{21}+1}{2^{2}}-6$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{21}\right)^{2}+2\sqrt{21}+1}{4}-6$$

The square of \sqrt{21} is 21.

$$\frac{21+2\sqrt{21}+1}{4}-6$$

Add 21 and 1 to get 22.

$$\frac{22+2\sqrt{21}}{4}-6$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 6 times \frac{4}{4}.

$$\frac{22+2\sqrt{21}}{4}-\frac{6\times 4}{4}$$

Since \frac{22+2\sqrt{21}}{4} and \frac{6\times 4}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{22+2\sqrt{21}-6\times 4}{4}$$

Do the multiplications in 22+2\sqrt{21}-6\times 4.

$$\frac{22+2\sqrt{21}-24}{4}$$

Do the calculations in 22+2\sqrt{21}-24.

$$\frac{-2+2\sqrt{21}}{4}$$