Solve 7+10+(28*(5-7+2)+47)-64+0/28-7+7*3

$$7+10+(28 \cdot (5-7+2)+47)-64+ \frac{ 0 }{ 28 } -7+7 \cdot 3$$

Evaluate

$$14$$

Solution Steps

Add 7 and 10 to get 17.

$$17+28\left(5-7+2\right)+47-64+\frac{0}{28}-7+7\times 3$$

Subtract 7 from 5 to get -2.

$$17+28\left(-2+2\right)+47-64+\frac{0}{28}-7+7\times 3$$

Add -2 and 2 to get 0.

$$17+28\times 0+47-64+\frac{0}{28}-7+7\times 3$$

Multiply 28 and 0 to get 0.

$$17+0+47-64+\frac{0}{28}-7+7\times 3$$

Add 17 and 0 to get 17.

$$17+47-64+\frac{0}{28}-7+7\times 3$$

Add 17 and 47 to get 64.

$$64-64+\frac{0}{28}-7+7\times 3$$

Subtract 64 from 64 to get 0.

$$\frac{0}{28}-7+7\times 3$$

Zero divided by any non-zero number gives zero.

$$0-7+7\times 3$$

Subtract 7 from 0 to get -7.

$$-7+7\times 3$$

Multiply 7 and 3 to get 21.

$$-7+21$$

Add -7 and 21 to get 14.

$$14$$

Factor

$$2\times 7$$