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-\frac{\left(\sqrt{5}+3\right)^{2}}{2^{2}}+2\times \frac{\sqrt{5}+3}{2}+3
To raise \frac{\sqrt{5}+3}{2} to a power, raise both numerator and denominator to the power and then divide.
-\frac{\left(\sqrt{5}+3\right)^{2}}{2^{2}}+\sqrt{5}+3+3
Cancel out 2 and 2.
-\frac{\left(\sqrt{5}+3\right)^{2}}{2^{2}}+\sqrt{5}+6
Add 3 and 3 to get 6.
-\frac{\left(\sqrt{5}+3\right)^{2}}{2^{2}}+\frac{\left(\sqrt{5}+6\right)\times 2^{2}}{2^{2}}
To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{5}+6 times \frac{2^{2}}{2^{2}}.
\frac{-\left(\sqrt{5}+3\right)^{2}+\left(\sqrt{5}+6\right)\times 2^{2}}{2^{2}}
Since -\frac{\left(\sqrt{5}+3\right)^{2}}{2^{2}} and \frac{\left(\sqrt{5}+6\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.
-\frac{\left(\sqrt{5}\right)^{2}+6\sqrt{5}+9}{2^{2}}+\sqrt{5}+6
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+3\right)^{2}.
-\frac{5+6\sqrt{5}+9}{2^{2}}+\sqrt{5}+6
The square of \sqrt{5} is 5.
-\frac{14+6\sqrt{5}}{2^{2}}+\sqrt{5}+6
Add 5 and 9 to get 14.
-\frac{14+6\sqrt{5}}{4}+\sqrt{5}+6
Calculate 2 to the power of 2 and get 4.
-\frac{14+6\sqrt{5}}{4}+\frac{4\left(\sqrt{5}+6\right)}{4}
To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{5}+6 times \frac{4}{4}.
\frac{-\left(14+6\sqrt{5}\right)+4\left(\sqrt{5}+6\right)}{4}
Since -\frac{14+6\sqrt{5}}{4} and \frac{4\left(\sqrt{5}+6\right)}{4} have the same denominator, add them by adding their numerators.
\frac{-14-6\sqrt{5}+4\sqrt{5}+24}{4}
Do the multiplications in -\left(14+6\sqrt{5}\right)+4\left(\sqrt{5}+6\right).
\frac{10-2\sqrt{5}}{4}
Do the calculations in -14-6\sqrt{5}+4\sqrt{5}+24.