Solve -(frac{sqrt{13}+3}{2})^2+sqrt{13}+3+3

$$- ( \frac { \sqrt { 13 } + 3 } { 2 } ) ^ { 2 } + \sqrt { 13 } + 3 + 3$$

Evaluate

$$\frac{1-\sqrt{13}}{2}\approx -1.302775638$$

Short Solution Steps

To raise \frac{\sqrt{13}+3}{2} to a power, raise both numerator and denominator to the power and then divide.

$$-\frac{\left(\sqrt{13}+3\right)^{2}}{2^{2}}+\sqrt{13}+3+3$$

Add 3 and 3 to get 6.

$$-\frac{\left(\sqrt{13}+3\right)^{2}}{2^{2}}+\sqrt{13}+6$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{13}+6 times \frac{2^{2}}{2^{2}}.

$$-\frac{\left(\sqrt{13}+3\right)^{2}}{2^{2}}+\frac{\left(\sqrt{13}+6\right)\times 2^{2}}{2^{2}}$$

Since -\frac{\left(\sqrt{13}+3\right)^{2}}{2^{2}} and \frac{\left(\sqrt{13}+6\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{-\left(\sqrt{13}+3\right)^{2}+\left(\sqrt{13}+6\right)\times 2^{2}}{2^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{13}+3\right)^{2}.

$$-\frac{\left(\sqrt{13}\right)^{2}+6\sqrt{13}+9}{2^{2}}+\sqrt{13}+6$$

The square of \sqrt{13} is 13.

$$-\frac{13+6\sqrt{13}+9}{2^{2}}+\sqrt{13}+6$$

Add 13 and 9 to get 22.

$$-\frac{22+6\sqrt{13}}{2^{2}}+\sqrt{13}+6$$

Calculate 2 to the power of 2 and get 4.

$$-\frac{22+6\sqrt{13}}{4}+\sqrt{13}+6$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{13}+6 times \frac{4}{4}.

$$-\frac{22+6\sqrt{13}}{4}+\frac{4\left(\sqrt{13}+6\right)}{4}$$

Since -\frac{22+6\sqrt{13}}{4} and \frac{4\left(\sqrt{13}+6\right)}{4} have the same denominator, add them by adding their numerators.

$$\frac{-\left(22+6\sqrt{13}\right)+4\left(\sqrt{13}+6\right)}{4}$$

Do the multiplications in -\left(22+6\sqrt{13}\right)+4\left(\sqrt{13}+6\right).

$$\frac{-22-6\sqrt{13}+4\sqrt{13}+24}{4}$$

Do the calculations in -22-6\sqrt{13}+4\sqrt{13}+24.

$$\frac{2-2\sqrt{13}}{4}$$

Factor

$$\frac{1 - \sqrt{13}}{2} = -1.3027756377319946$$