Solve 2-(sqrt{6}+6sqrt{7})^2

$$2 - ( \sqrt { 6 } + 6 \sqrt { 7 } ) ^ { 2 }$$

Evaluate

$$-12\sqrt{42}-256\approx -333.768888381$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+6\sqrt{7}\right)^{2}.

$$2-\left(\left(\sqrt{6}\right)^{2}+12\sqrt{6}\sqrt{7}+36\left(\sqrt{7}\right)^{2}\right)$$

The square of \sqrt{6} is 6.

$$2-\left(6+12\sqrt{6}\sqrt{7}+36\left(\sqrt{7}\right)^{2}\right)$$

To multiply \sqrt{6} and \sqrt{7}, multiply the numbers under the square root.

$$2-\left(6+12\sqrt{42}+36\left(\sqrt{7}\right)^{2}\right)$$

The square of \sqrt{7} is 7.

$$2-\left(6+12\sqrt{42}+36\times 7\right)$$

Multiply 36 and 7 to get 252.

$$2-\left(6+12\sqrt{42}+252\right)$$

Add 6 and 252 to get 258.

$$2-\left(258+12\sqrt{42}\right)$$

To find the opposite of 258+12\sqrt{42}, find the opposite of each term.

$$2-258-12\sqrt{42}$$

Subtract 258 from 2 to get -256.

$$-256-12\sqrt{42}$$

Expand

$$-12\sqrt{42}-256$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+6\sqrt{7}\right)^{2}.

$$2-\left(\left(\sqrt{6}\right)^{2}+12\sqrt{6}\sqrt{7}+36\left(\sqrt{7}\right)^{2}\right)$$

The square of \sqrt{6} is 6.

$$2-\left(6+12\sqrt{6}\sqrt{7}+36\left(\sqrt{7}\right)^{2}\right)$$

To multiply \sqrt{6} and \sqrt{7}, multiply the numbers under the square root.

$$2-\left(6+12\sqrt{42}+36\left(\sqrt{7}\right)^{2}\right)$$

The square of \sqrt{7} is 7.

$$2-\left(6+12\sqrt{42}+36\times 7\right)$$

Multiply 36 and 7 to get 252.

$$2-\left(6+12\sqrt{42}+252\right)$$

Add 6 and 252 to get 258.

$$2-\left(258+12\sqrt{42}\right)$$

To find the opposite of 258+12\sqrt{42}, find the opposite of each term.

$$2-258-12\sqrt{42}$$

Subtract 258 from 2 to get -256.

$$-256-12\sqrt{42}$$