Solve (2sqrt{3})^2+(3-2sqrt{2})*(5+sqrt{8})=

$$( 2 \sqrt { 3 } ) ^ { 2 } + ( 3 - 2 \sqrt { 2 } ) \cdot ( 5 + \sqrt { 8 } ) =$$

Evaluate

$$19-4\sqrt{2}\approx 13.343145751$$

Solution Steps

Expand \left(2\sqrt{3}\right)^{2}.

$$2^{2}\left(\sqrt{3}\right)^{2}+\left(3-2\sqrt{2}\right)\left(5+\sqrt{8}\right)$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3}\right)^{2}+\left(3-2\sqrt{2}\right)\left(5+\sqrt{8}\right)$$

The square of \sqrt{3} is 3.

$$4\times 3+\left(3-2\sqrt{2}\right)\left(5+\sqrt{8}\right)$$

Multiply 4 and 3 to get 12.

$$12+\left(3-2\sqrt{2}\right)\left(5+\sqrt{8}\right)$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$12+\left(3-2\sqrt{2}\right)\left(5+2\sqrt{2}\right)$$

Use the distributive property to multiply 3-2\sqrt{2} by 5+2\sqrt{2} and combine like terms.

$$12+15-4\sqrt{2}-4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$12+15-4\sqrt{2}-4\times 2$$

Multiply -4 and 2 to get -8.

$$12+15-4\sqrt{2}-8$$

Subtract 8 from 15 to get 7.

$$12+7-4\sqrt{2}$$

Add 12 and 7 to get 19.

$$19-4\sqrt{2}$$