Solve 4^2*5-6*sqrt{25}+2^2*8^0=

$$4 ^ { 2 } \cdot 5 - 6 \cdot \sqrt { 25 } + 2 ^ { 2 } \cdot 8 ^ { 0 } =$$

Evaluate

$$54$$

Solution Steps

Calculate 4 to the power of 2 and get 16.

$$16\times 5-6\sqrt{25}+2^{2}\times 8^{0}$$

Multiply 16 and 5 to get 80.

$$80-6\sqrt{25}+2^{2}\times 8^{0}$$

Calculate the square root of 25 and get 5.

$$80-6\times 5+2^{2}\times 8^{0}$$

Multiply -6 and 5 to get -30.

$$80-30+2^{2}\times 8^{0}$$

Subtract 30 from 80 to get 50.

$$50+2^{2}\times 8^{0}$$

Calculate 2 to the power of 2 and get 4.

$$50+4\times 8^{0}$$

Calculate 8 to the power of 0 and get 1.

$$50+4\times 1$$

Multiply 4 and 1 to get 4.

$$50+4$$

Add 50 and 4 to get 54.

$$54$$

Factor

$$2\times 3^{3}$$