Solve (6^2-7*5+2)+31-sqrt{(2^3+6*7)}:10-6^0

$$( 6 ^ { 2 } - 7 \times 5 + 2 ) + 31 - \sqrt { ( 2 ^ { 3 } + 6 \times 7 ) } : 10 - 6 ^ { 0 }$$

Evaluate

$$-\frac{\sqrt{2}}{2}+33\approx 32.292893219$$

Solution Steps

Calculate 6 to the power of 2 and get 36.

$$36-7\times 5+2+31-\frac{\sqrt{2^{3}+6\times 7}}{10}-6^{0}$$

Multiply 7 and 5 to get 35.

$$36-35+2+31-\frac{\sqrt{2^{3}+6\times 7}}{10}-6^{0}$$

Subtract 35 from 36 to get 1.

$$1+2+31-\frac{\sqrt{2^{3}+6\times 7}}{10}-6^{0}$$

Add 1 and 2 to get 3.

$$3+31-\frac{\sqrt{2^{3}+6\times 7}}{10}-6^{0}$$

Add 3 and 31 to get 34.

$$34-\frac{\sqrt{2^{3}+6\times 7}}{10}-6^{0}$$

Calculate 2 to the power of 3 and get 8.

$$34-\frac{\sqrt{8+6\times 7}}{10}-6^{0}$$

Multiply 6 and 7 to get 42.

$$34-\frac{\sqrt{8+42}}{10}-6^{0}$$

Add 8 and 42 to get 50.

$$34-\frac{\sqrt{50}}{10}-6^{0}$$

Factor 50=5^{2}\times 2. Rewrite the square root of the product \sqrt{5^{2}\times 2} as the product of square roots \sqrt{5^{2}}\sqrt{2}. Take the square root of 5^{2}.

$$34-\frac{5\sqrt{2}}{10}-6^{0}$$

Divide 5\sqrt{2} by 10 to get \frac{1}{2}\sqrt{2}.

$$34-\frac{1}{2}\sqrt{2}-6^{0}$$

Calculate 6 to the power of 0 and get 1.

$$34-\frac{1}{2}\sqrt{2}-1$$

Subtract 1 from 34 to get 33.

$$33-\frac{1}{2}\sqrt{2}$$

Factor

$$\frac{66 - \sqrt{2}}{2} = 32.292893218813454$$