Solve 20+25+36-6-83*5+87/7

$$20+25+36-6-83 \cdot 5+ \frac{ 87 }{ 7 }$$

Evaluate

$$-\frac{2293}{7}\approx -327.571428571$$

Solution Steps

Add 20 and 25 to get 45.

$$45+36-6-83\times 5+\frac{87}{7}$$

Add 45 and 36 to get 81.

$$81-6-83\times 5+\frac{87}{7}$$

Subtract 6 from 81 to get 75.

$$75-83\times 5+\frac{87}{7}$$

Multiply 83 and 5 to get 415.

$$75-415+\frac{87}{7}$$

Subtract 415 from 75 to get -340.

$$-340+\frac{87}{7}$$

Convert -340 to fraction -\frac{2380}{7}.

$$-\frac{2380}{7}+\frac{87}{7}$$

Since -\frac{2380}{7} and \frac{87}{7} have the same denominator, add them by adding their numerators.

$$\frac{-2380+87}{7}$$

Add -2380 and 87 to get -2293.

$$-\frac{2293}{7}$$

Factor

$$-\frac{2293}{7} = -327\frac{4}{7} = -327.57142857142856$$