Solve 2(2-sqrt{3})^2+6(2-sqrt{3})

$$2 ( 2 - \sqrt { 3 } ) ^ { 2 } + 6 ( 2 - \sqrt { 3 } )$$

Evaluate

$$26-14\sqrt{3}\approx 1.751288694$$

Solution Steps

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

$$2\left(4-4\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)+6\left(2-\sqrt{3}\right)$$

The square of \sqrt{3} is 3.

$$2\left(4-4\sqrt{3}+3\right)+6\left(2-\sqrt{3}\right)$$

Add 4 and 3 to get 7.

$$2\left(7-4\sqrt{3}\right)+6\left(2-\sqrt{3}\right)$$

Use the distributive property to multiply 2 by 7-4\sqrt{3}.

$$14-8\sqrt{3}+6\left(2-\sqrt{3}\right)$$

Use the distributive property to multiply 6 by 2-\sqrt{3}.

$$14-8\sqrt{3}+12-6\sqrt{3}$$

Add 14 and 12 to get 26.

$$26-8\sqrt{3}-6\sqrt{3}$$

Combine -8\sqrt{3} and -6\sqrt{3} to get -14\sqrt{3}.

$$26-14\sqrt{3}$$

Factor

$$2 {(13 - 7 \sqrt{3})} = 1.751288694$$