Solve 3{left(frac{sqrt{2505}}{6}-15/2right)}^2

$$3 { \left( \frac{ \sqrt{ 2505 } }{ 6 } - \frac{ 15 }{ 2 } \right) }^{ 2 }$$

Evaluate

$$\frac{755-15\sqrt{2505}}{2}\approx 2.125187313$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6 and 2 is 6. Multiply \frac{15}{2} times \frac{3}{3}.

$$3\left(\frac{\sqrt{2505}}{6}-\frac{15\times 3}{6}\right)^{2}$$

Since \frac{\sqrt{2505}}{6} and \frac{15\times 3}{6} have the same denominator, subtract them by subtracting their numerators.

$$3\times \left(\frac{\sqrt{2505}-15\times 3}{6}\right)^{2}$$

Do the multiplications in \sqrt{2505}-15\times 3.

$$3\times \left(\frac{\sqrt{2505}-45}{6}\right)^{2}$$

To raise \frac{\sqrt{2505}-45}{6} to a power, raise both numerator and denominator to the power and then divide.

$$3\times \frac{\left(\sqrt{2505}-45\right)^{2}}{6^{2}}$$

Express 3\times \frac{\left(\sqrt{2505}-45\right)^{2}}{6^{2}} as a single fraction.

$$\frac{3\left(\sqrt{2505}-45\right)^{2}}{6^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2505}-45\right)^{2}.

$$\frac{3\left(\left(\sqrt{2505}\right)^{2}-90\sqrt{2505}+2025\right)}{6^{2}}$$

The square of \sqrt{2505} is 2505.

$$\frac{3\left(2505-90\sqrt{2505}+2025\right)}{6^{2}}$$

Add 2505 and 2025 to get 4530.

$$\frac{3\left(4530-90\sqrt{2505}\right)}{6^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{3\left(4530-90\sqrt{2505}\right)}{36}$$

Divide 3\left(4530-90\sqrt{2505}\right) by 36 to get \frac{1}{12}\left(4530-90\sqrt{2505}\right).

$$\frac{1}{12}\left(4530-90\sqrt{2505}\right)$$

Use the distributive property to multiply \frac{1}{12} by 4530-90\sqrt{2505}.

$$\frac{755}{2}-\frac{15}{2}\sqrt{2505}$$

Expand

$$\frac{755 - 15 \sqrt{2505}}{2} = 2.125187312734056$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6 and 2 is 6. Multiply \frac{15}{2} times \frac{3}{3}.

$$3\left(\frac{\sqrt{2505}}{6}-\frac{15\times 3}{6}\right)^{2}$$

Since \frac{\sqrt{2505}}{6} and \frac{15\times 3}{6} have the same denominator, subtract them by subtracting their numerators.

$$3\times \left(\frac{\sqrt{2505}-15\times 3}{6}\right)^{2}$$

Do the multiplications in \sqrt{2505}-15\times 3.

$$3\times \left(\frac{\sqrt{2505}-45}{6}\right)^{2}$$

To raise \frac{\sqrt{2505}-45}{6} to a power, raise both numerator and denominator to the power and then divide.

$$3\times \frac{\left(\sqrt{2505}-45\right)^{2}}{6^{2}}$$

Express 3\times \frac{\left(\sqrt{2505}-45\right)^{2}}{6^{2}} as a single fraction.

$$\frac{3\left(\sqrt{2505}-45\right)^{2}}{6^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2505}-45\right)^{2}.

$$\frac{3\left(\left(\sqrt{2505}\right)^{2}-90\sqrt{2505}+2025\right)}{6^{2}}$$

The square of \sqrt{2505} is 2505.

$$\frac{3\left(2505-90\sqrt{2505}+2025\right)}{6^{2}}$$

Add 2505 and 2025 to get 4530.

$$\frac{3\left(4530-90\sqrt{2505}\right)}{6^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{3\left(4530-90\sqrt{2505}\right)}{36}$$

Divide 3\left(4530-90\sqrt{2505}\right) by 36 to get \frac{1}{12}\left(4530-90\sqrt{2505}\right).

$$\frac{1}{12}\left(4530-90\sqrt{2505}\right)$$

Use the distributive property to multiply \frac{1}{12} by 4530-90\sqrt{2505}.

$$\frac{755}{2}-\frac{15}{2}\sqrt{2505}$$