Solve sqrt{4^2+3^2}-2*(3^2-2^2)

$$\sqrt { 4 ^ { 2 } + 3 ^ { 2 } } - 2 \cdot ( 3 ^ { 2 } - 2 ^ { 2 } )$$

Evaluate

$$-5$$

Solution Steps

Calculate 4 to the power of 2 and get 16.

$$\sqrt{16+3^{2}}-2\left(3^{2}-2^{2}\right)$$

Calculate 3 to the power of 2 and get 9.

$$\sqrt{16+9}-2\left(3^{2}-2^{2}\right)$$

Add 16 and 9 to get 25.

$$\sqrt{25}-2\left(3^{2}-2^{2}\right)$$

Calculate the square root of 25 and get 5.

$$5-2\left(3^{2}-2^{2}\right)$$

Calculate 3 to the power of 2 and get 9.

$$5-2\left(9-2^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$5-2\left(9-4\right)$$

Subtract 4 from 9 to get 5.

$$5-2\times 5$$

Multiply 2 and 5 to get 10.

$$5-10$$

Subtract 10 from 5 to get -5.

$$-5$$

Factor

$$-5$$