Solve {left(frac{-sqrt{13}-1}{2}right)}^2-frac{sqrt{13}-1}{2}+2019

$${ \left( \frac{ - \sqrt{ 13 } -1 }{ 2 } \right) }^{ 2 } - \frac{ \sqrt{ 13 } -1 }{ 2 } +2019$$

Evaluate

$$2023$$

Short Solution Steps

To raise \frac{-\sqrt{13}-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-\sqrt{13}-1\right)^{2}}{2^{2}}-\frac{\sqrt{13}-1}{2}+2019$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2^{2} and 2 is 4. Multiply \frac{\sqrt{13}-1}{2} times \frac{2}{2}.

$$\frac{\left(-\sqrt{13}-1\right)^{2}}{4}-\frac{2\left(\sqrt{13}-1\right)}{4}+2019$$

Since \frac{\left(-\sqrt{13}-1\right)^{2}}{4} and \frac{2\left(\sqrt{13}-1\right)}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(-\sqrt{13}-1\right)^{2}-2\left(\sqrt{13}-1\right)}{4}+2019$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2019 times \frac{2^{2}}{2^{2}}.

$$\frac{\left(-\sqrt{13}-1\right)^{2}}{2^{2}}-\frac{\sqrt{13}-1}{2}+\frac{2019\times 2^{2}}{2^{2}}$$

Since \frac{\left(-\sqrt{13}-1\right)^{2}}{2^{2}} and \frac{2019\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(-\sqrt{13}-1\right)^{2}+2019\times 2^{2}}{2^{2}}-\frac{\sqrt{13}-1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2019 times \frac{2}{2}.

$$\frac{\left(-\sqrt{13}-1\right)^{2}}{2^{2}}-\frac{\sqrt{13}-1}{2}+\frac{2019\times 2}{2}$$

Since -\frac{\sqrt{13}-1}{2} and \frac{2019\times 2}{2} have the same denominator, add them by adding their numerators.

$$\frac{\left(-\sqrt{13}-1\right)^{2}}{2^{2}}+\frac{-\left(\sqrt{13}-1\right)+2019\times 2}{2}$$

Do the multiplications in -\left(\sqrt{13}-1\right)+2019\times 2.

$$\frac{\left(-\sqrt{13}-1\right)^{2}}{2^{2}}+\frac{-\sqrt{13}+1+4038}{2}$$

Do the calculations in -\sqrt{13}+1+4038.

$$\frac{\left(-\sqrt{13}-1\right)^{2}}{2^{2}}+\frac{-\sqrt{13}+4039}{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-\sqrt{13}-1\right)^{2}.

$$\frac{\left(-\sqrt{13}\right)^{2}-2\left(-\sqrt{13}\right)+1}{2^{2}}+\frac{-\sqrt{13}+4039}{2}$$

Calculate -\sqrt{13} to the power of 2 and get \left(\sqrt{13}\right)^{2}.

$$\frac{\left(\sqrt{13}\right)^{2}-2\left(-\sqrt{13}\right)+1}{2^{2}}+\frac{-\sqrt{13}+4039}{2}$$

Multiply -2 and -1 to get 2.

$$\frac{\left(\sqrt{13}\right)^{2}+2\sqrt{13}+1}{2^{2}}+\frac{-\sqrt{13}+4039}{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{13}\right)^{2}+2\sqrt{13}+1}{4}+\frac{-\sqrt{13}+4039}{2}$$

The square of \sqrt{13} is 13.

$$\frac{13+2\sqrt{13}+1}{4}+\frac{-\sqrt{13}+4039}{2}$$

Add 13 and 1 to get 14.

$$\frac{14+2\sqrt{13}}{4}+\frac{-\sqrt{13}+4039}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2 is 4. Multiply \frac{-\sqrt{13}+4039}{2} times \frac{2}{2}.

$$\frac{14+2\sqrt{13}}{4}+\frac{2\left(-\sqrt{13}+4039\right)}{4}$$

Since \frac{14+2\sqrt{13}}{4} and \frac{2\left(-\sqrt{13}+4039\right)}{4} have the same denominator, add them by adding their numerators.

$$\frac{14+2\sqrt{13}+2\left(-\sqrt{13}+4039\right)}{4}$$

Do the multiplications in 14+2\sqrt{13}+2\left(-\sqrt{13}+4039\right).

$$\frac{14+2\sqrt{13}-2\sqrt{13}+8078}{4}$$

Do the calculations in 14+2\sqrt{13}-2\sqrt{13}+8078.

$$\frac{8092}{4}$$

Factor

$$7\times 17^{2}$$