Solve sqrt12^2:(-sqrt{5})^2+(sqrt{3})^2*sqrt{2^2}

$$\sqrt { 12 } ^ { 2 } : ( - \sqrt { 5 } ) ^ { 2 } + ( \sqrt { 3 } ) ^ { 2 } \times \sqrt { 2 ^ { 2 } }$$

Evaluate

$$\frac{42}{5}=8.4$$

Solution Steps

The square of \sqrt{12} is 12.

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

Calculate -\sqrt{5} to the power of 2 and get \left(\sqrt{5}\right)^{2}.

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

The square of \sqrt{5} is 5.

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

The square of \sqrt{3} is 3.

$$\frac{12}{5}+3\sqrt{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{12}{5}+3\sqrt{4}$$

Calculate the square root of 4 and get 2.

$$\frac{12}{5}+3\times 2$$

Multiply 3 and 2 to get 6.

$$\frac{12}{5}+6$$

Add \frac{12}{5} and 6 to get \frac{42}{5}.

$$\frac{42}{5}$$

Factor

$$\frac{2 \cdot 3 \cdot 7}{5} = 8\frac{2}{5} = 8.4$$