Solve [(10-5+3)^2-3*10]:(4^2+1)+13

$$[ ( 10 - 5 + 3 ) ^ { 2 } - 3 \cdot 10 ] : ( 4 ^ { 2 } + 1 ) + 13$$

Evaluate

$$15$$

Solution Steps

Subtract 5 from 10 to get 5.

$$\frac{\left(5+3\right)^{2}-3\times 10}{4^{2}+1}+13$$

Add 5 and 3 to get 8.

$$\frac{8^{2}-3\times 10}{4^{2}+1}+13$$

Calculate 8 to the power of 2 and get 64.

$$\frac{64-3\times 10}{4^{2}+1}+13$$

Multiply 3 and 10 to get 30.

$$\frac{64-30}{4^{2}+1}+13$$

Subtract 30 from 64 to get 34.

$$\frac{34}{4^{2}+1}+13$$

Calculate 4 to the power of 2 and get 16.

$$\frac{34}{16+1}+13$$

Add 16 and 1 to get 17.

$$\frac{34}{17}+13$$

Divide 34 by 17 to get 2.

$$2+13$$

Add 2 and 13 to get 15.

$$15$$

Factor

$$3\times 5$$