Solve {left(frac{sqrt{33}+3}{2}right)}^2-sqrt{33}-3-6

$${ \left( \frac{ \sqrt{ 33 } +3 }{ 2 } \right) }^{ 2 } - \sqrt{ 33 } -3-6$$

Evaluate

$$\frac{\sqrt{33}+3}{2}\approx 4.372281323$$

Short Solution Steps

To raise \frac{\sqrt{33}+3}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{33}+3\right)^{2}}{2^{2}}-\sqrt{33}-3-6$$

Subtract 6 from -3 to get -9.

$$\frac{\left(\sqrt{33}+3\right)^{2}}{2^{2}}-\sqrt{33}-9$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -\sqrt{33}-9 times \frac{2^{2}}{2^{2}}.

$$\frac{\left(\sqrt{33}+3\right)^{2}}{2^{2}}+\frac{\left(-\sqrt{33}-9\right)\times 2^{2}}{2^{2}}$$

Since \frac{\left(\sqrt{33}+3\right)^{2}}{2^{2}} and \frac{\left(-\sqrt{33}-9\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(\sqrt{33}+3\right)^{2}+\left(-\sqrt{33}-9\right)\times 2^{2}}{2^{2}}$$

Do the multiplications in \left(\sqrt{33}+3\right)^{2}+\left(-\sqrt{33}-9\right)\times 2^{2}.

$$\frac{\left(\sqrt{33}\right)^{2}+6\sqrt{33}+9-4\sqrt{33}-36}{2^{2}}$$

Do the calculations in \left(\sqrt{33}\right)^{2}+6\sqrt{33}+9-4\sqrt{33}-36.

$$\frac{6+2\sqrt{33}}{2^{2}}$$

Expand 2^{2}.

$$\frac{6+2\sqrt{33}}{4}$$

Factor

$$\frac{\sqrt{33} + 3}{2} = 4.372281323269014$$