Solve -2{left(1+sqrt{3}right)}^2-3left(1+sqrt{3}right)-2

$$-2 { \left(1+ \sqrt{ 3 } \right) }^{ 2 } -3 \left( 1+ \sqrt{ 3 } \right) -2$$

Evaluate

$$-7\sqrt{3}-13\approx -25.124355653$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{3}\right)^{2}.

$$-2\left(1+2\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)-3\left(1+\sqrt{3}\right)-2$$

The square of \sqrt{3} is 3.

$$-2\left(1+2\sqrt{3}+3\right)-3\left(1+\sqrt{3}\right)-2$$

Add 1 and 3 to get 4.

$$-2\left(4+2\sqrt{3}\right)-3\left(1+\sqrt{3}\right)-2$$

Use the distributive property to multiply -2 by 4+2\sqrt{3}.

$$-8-4\sqrt{3}-3\left(1+\sqrt{3}\right)-2$$

Use the distributive property to multiply -3 by 1+\sqrt{3}.

$$-8-4\sqrt{3}-3-3\sqrt{3}-2$$

Subtract 3 from -8 to get -11.

$$-11-4\sqrt{3}-3\sqrt{3}-2$$

Combine -4\sqrt{3} and -3\sqrt{3} to get -7\sqrt{3}.

$$-11-7\sqrt{3}-2$$

Subtract 2 from -11 to get -13.

$$-13-7\sqrt{3}$$