Solve sqrt{3*4+2*3*4/3^2+4^2}=

$$\sqrt { \frac { 3 \cdot 4 + 2 \cdot 3 \cdot 4 } { 3 ^ { 2 } + 4 ^ { 2 } } } =$$

Evaluate

$$\frac{6}{5}=1.2$$

Solution Steps

Multiply 3 and 4 to get 12. Multiply 2 and 3 to get 6.

$$\sqrt{\frac{12+6\times 4}{3^{2}+4^{2}}}$$

Multiply 6 and 4 to get 24.

$$\sqrt{\frac{12+24}{3^{2}+4^{2}}}$$

Add 12 and 24 to get 36.

$$\sqrt{\frac{36}{3^{2}+4^{2}}}$$

Calculate 3 to the power of 2 and get 9.

$$\sqrt{\frac{36}{9+4^{2}}}$$

Calculate 4 to the power of 2 and get 16.

$$\sqrt{\frac{36}{9+16}}$$

Add 9 and 16 to get 25.

$$\sqrt{\frac{36}{25}}$$

Rewrite the square root of the division \frac{36}{25} as the division of square roots \frac{\sqrt{36}}{\sqrt{25}}. Take the square root of both numerator and denominator.

$$\frac{6}{5}$$

Factor

$$\frac{2 \cdot 3}{5} = 1\frac{1}{5} = 1.2$$