Solve (7^2-4^2-3^2)+(sqrt{25}*sqrt{16}-8^2)

$$( 7 ^ { 2 } - 4 ^ { 2 } - 3 ^ { 2 } ) + ( \sqrt { 25 } \times \sqrt { 16 } - 8 ^ { 2 } )$$

Evaluate

$$-20$$

Solution Steps

Calculate 7 to the power of 2 and get 49.

$$49-4^{2}-3^{2}+\sqrt{25}\sqrt{16}-8^{2}$$

Calculate 4 to the power of 2 and get 16.

$$49-16-3^{2}+\sqrt{25}\sqrt{16}-8^{2}$$

Subtract 16 from 49 to get 33.

$$33-3^{2}+\sqrt{25}\sqrt{16}-8^{2}$$

Calculate 3 to the power of 2 and get 9.

$$33-9+\sqrt{25}\sqrt{16}-8^{2}$$

Subtract 9 from 33 to get 24.

$$24+\sqrt{25}\sqrt{16}-8^{2}$$

Calculate the square root of 25 and get 5.

$$24+5\sqrt{16}-8^{2}$$

Calculate the square root of 16 and get 4.

$$24+5\times 4-8^{2}$$

Multiply 5 and 4 to get 20.

$$24+20-8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$24+20-64$$

Subtract 64 from 20 to get -44.

$$24-44$$

Subtract 44 from 24 to get -20.

$$-20$$

Factor

$$-20$$