Solve sqrt{(11*32)+7^2-1^2}+4^2+3^2=

$$\sqrt { ( 11 \times 32 ) + 7 ^ { 2 } - 1 ^ { 2 } } + 4 ^ { 2 } + 3 ^ { 2 } =$$

Evaluate

$$45$$

Solution Steps

Multiply 11 and 32 to get 352.

$$\sqrt{352+7^{2}-1^{2}}+4^{2}+3^{2}$$

Calculate 7 to the power of 2 and get 49.

$$\sqrt{352+49-1^{2}}+4^{2}+3^{2}$$

Add 352 and 49 to get 401.

$$\sqrt{401-1^{2}}+4^{2}+3^{2}$$

Calculate 1 to the power of 2 and get 1.

$$\sqrt{401-1}+4^{2}+3^{2}$$

Subtract 1 from 401 to get 400.

$$\sqrt{400}+4^{2}+3^{2}$$

Calculate the square root of 400 and get 20.

$$20+4^{2}+3^{2}$$

Calculate 4 to the power of 2 and get 16.

$$20+16+3^{2}$$

Add 20 and 16 to get 36.

$$36+3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$36+9$$

Add 36 and 9 to get 45.

$$45$$

Factor

$$3^{2}\times 5$$