Solve (3+2sqrt{5})^2-sqrt{720}-18

$$( 3 + 2 \sqrt { 5 } ) ^ { 2 } - \sqrt { 720 } - 18$$

Evaluate

$$11$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+2\sqrt{5}\right)^{2}.

$$9+12\sqrt{5}+4\left(\sqrt{5}\right)^{2}-\sqrt{720}-18$$

The square of \sqrt{5} is 5.

$$9+12\sqrt{5}+4\times 5-\sqrt{720}-18$$

Multiply 4 and 5 to get 20.

$$9+12\sqrt{5}+20-\sqrt{720}-18$$

Add 9 and 20 to get 29.

$$29+12\sqrt{5}-\sqrt{720}-18$$

Factor 720=12^{2}\times 5. Rewrite the square root of the product \sqrt{12^{2}\times 5} as the product of square roots \sqrt{12^{2}}\sqrt{5}. Take the square root of 12^{2}.

$$29+12\sqrt{5}-12\sqrt{5}-18$$

Combine 12\sqrt{5} and -12\sqrt{5} to get 0.

$$29-18$$

Subtract 18 from 29 to get 11.

$$11$$

Factor

$$11$$