Solve (2^2*11-3^2*2):2+4^2-19

$$( 2 ^ { 2 } \cdot 11 - 3 ^ { 2 } \cdot 2 ) : 2 + 4 ^ { 2 } - 19$$

Evaluate

$$10$$

Solution Steps

Calculate 2 to the power of 2 and get 4.

$$\frac{4\times 11-3^{2}\times 2}{2}+4^{2}-19$$

Multiply 4 and 11 to get 44.

$$\frac{44-3^{2}\times 2}{2}+4^{2}-19$$

Calculate 3 to the power of 2 and get 9.

$$\frac{44-9\times 2}{2}+4^{2}-19$$

Multiply 9 and 2 to get 18.

$$\frac{44-18}{2}+4^{2}-19$$

Subtract 18 from 44 to get 26.

$$\frac{26}{2}+4^{2}-19$$

Divide 26 by 2 to get 13.

$$13+4^{2}-19$$

Calculate 4 to the power of 2 and get 16.

$$13+16-19$$

Add 13 and 16 to get 29.

$$29-19$$

Subtract 19 from 29 to get 10.

$$10$$

Factor

$$2\times 5$$