Solve 0,25*(3*2^3+4^2)-2,1^2

$$0,25 \cdot ( 3 \cdot 2 ^ { 3 } + 4 ^ { 2 } ) - 2,1 ^ { 2 }$$

Evaluate

$$8$$

Solution Steps

Calculate 2 to the power of 3 and get 8.

$$0,25\left(3\times 8+4^{2}\right)-2\times 1^{2}$$

Multiply 3 and 8 to get 24.

$$0,25\left(24+4^{2}\right)-2\times 1^{2}$$

Calculate 4 to the power of 2 and get 16.

$$0,25\left(24+16\right)-2\times 1^{2}$$

Add 24 and 16 to get 40.

$$0,25\times 40-2\times 1^{2}$$

Multiply 0,25 and 40 to get 10.

$$10-2\times 1^{2}$$

Calculate 1 to the power of 2 and get 1.

$$10-2\times 1$$

Multiply 2 and 1 to get 2.

$$10-2$$

Subtract 2 from 10 to get 8.

$$8$$

Factor

$$2^{3}$$