Solve 6frac{sqrt{21}}{sqrt{3^2}}-{left(2sqrt{7}+3sqrt{3}right)}^2

$$6 \frac{ \sqrt{ 21 } }{ \sqrt{ { 3 }^{ 2 } } } - { \left(2 \sqrt{ 7 } +3 \sqrt{ 3 } \right) }^{ 2 }$$

Evaluate

$$-10\sqrt{21}-55\approx -100.82575695$$

Solution Steps

Calculate 3 to the power of 2 and get 9.

$$6\times \frac{\sqrt{21}}{\sqrt{9}}-\left(2\sqrt{7}+3\sqrt{3}\right)^{2}$$

Calculate the square root of 9 and get 3.

$$6\times \frac{\sqrt{21}}{3}-\left(2\sqrt{7}+3\sqrt{3}\right)^{2}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$2\sqrt{21}-\left(2\sqrt{7}+3\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2\sqrt{7}+3\sqrt{3}\right)^{2}.

$$2\sqrt{21}-\left(4\left(\sqrt{7}\right)^{2}+12\sqrt{7}\sqrt{3}+9\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{7} is 7.

$$2\sqrt{21}-\left(4\times 7+12\sqrt{7}\sqrt{3}+9\left(\sqrt{3}\right)^{2}\right)$$

Multiply 4 and 7 to get 28.

$$2\sqrt{21}-\left(28+12\sqrt{7}\sqrt{3}+9\left(\sqrt{3}\right)^{2}\right)$$

To multiply \sqrt{7} and \sqrt{3}, multiply the numbers under the square root.

$$2\sqrt{21}-\left(28+12\sqrt{21}+9\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$2\sqrt{21}-\left(28+12\sqrt{21}+9\times 3\right)$$

Multiply 9 and 3 to get 27.

$$2\sqrt{21}-\left(28+12\sqrt{21}+27\right)$$

Add 28 and 27 to get 55.

$$2\sqrt{21}-\left(55+12\sqrt{21}\right)$$

To find the opposite of 55+12\sqrt{21}, find the opposite of each term.

$$2\sqrt{21}-55-12\sqrt{21}$$

Combine 2\sqrt{21} and -12\sqrt{21} to get -10\sqrt{21}.

$$-10\sqrt{21}-55$$