Solve 4^2div2^2-5*(-2)^2+5^2=

$$4 ^ { 2 } \div 2 ^ { 2 } - 5 \cdot ( - 2 ) ^ { 2 } + 5 ^ { 2 } =$$

Evaluate

$$9$$

Solution Steps

Calculate 4 to the power of 2 and get 16.

$$\frac{16}{2^{2}}-5\left(-2\right)^{2}+5^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{16}{4}-5\left(-2\right)^{2}+5^{2}$$

Divide 16 by 4 to get 4.

$$4-5\left(-2\right)^{2}+5^{2}$$

Calculate -2 to the power of 2 and get 4.

$$4-5\times 4+5^{2}$$

Multiply 5 and 4 to get 20.

$$4-20+5^{2}$$

Subtract 20 from 4 to get -16.

$$-16+5^{2}$$

Calculate 5 to the power of 2 and get 25.

$$-16+25$$

Add -16 and 25 to get 9.

$$9$$

Factor

$$3^{2}$$