Solve 7-84*01+3*(-4*025+3^2)+41+2

$$7 - 84 \cdot 01 + 3 \cdot ( - 4 \cdot 025 + 3 ^ { 2 } ) + 41 + 2$$

Evaluate

$$77$$

Solution Steps

Multiply 84 and 0 to get 0.

$$7-0\times 1+3\left(-4\times 0\times 25+3^{2}\right)+41+2$$

Multiply 0 and 1 to get 0.

$$7-0+3\left(-4\times 0\times 25+3^{2}\right)+41+2$$

Subtract 0 from 7 to get 7.

$$7+3\left(-4\times 0\times 25+3^{2}\right)+41+2$$

Multiply -4 and 0 to get 0.

$$7+3\left(0\times 25+3^{2}\right)+41+2$$

Multiply 0 and 25 to get 0.

$$7+3\left(0+3^{2}\right)+41+2$$

Calculate 3 to the power of 2 and get 9.

$$7+3\left(0+9\right)+41+2$$

Add 0 and 9 to get 9.

$$7+3\times 9+41+2$$

Multiply 3 and 9 to get 27.

$$7+27+41+2$$

Add 7 and 27 to get 34.

$$34+41+2$$

Add 34 and 41 to get 75.

$$75+2$$

Add 75 and 2 to get 77.

$$77$$

Factor

$$7\times 11$$