Solve sqrt{13^2-10^2+3*4+7^2-3^2+(96div{2}^4}

$$\sqrt{ { 13 }^{ 2 } - { 10 }^{ 2 } +3 \times 4+ { 7 }^{ 2 } - { 3 }^{ 2 } +(96 \div { 2 }^{ 4 } }$$

Evaluate

$$\sqrt{127}\approx 11.26942767$$

Solution Steps

Calculate 13 to the power of 2 and get 169.

$$\sqrt{169-10^{2}+3\times 4+7^{2}-3^{2}+\frac{96}{2^{4}}}$$

Calculate 10 to the power of 2 and get 100.

$$\sqrt{169-100+3\times 4+7^{2}-3^{2}+\frac{96}{2^{4}}}$$

Subtract 100 from 169 to get 69.

$$\sqrt{69+3\times 4+7^{2}-3^{2}+\frac{96}{2^{4}}}$$

Multiply 3 and 4 to get 12.

$$\sqrt{69+12+7^{2}-3^{2}+\frac{96}{2^{4}}}$$

Add 69 and 12 to get 81.

$$\sqrt{81+7^{2}-3^{2}+\frac{96}{2^{4}}}$$

Calculate 7 to the power of 2 and get 49.

$$\sqrt{81+49-3^{2}+\frac{96}{2^{4}}}$$

Add 81 and 49 to get 130.

$$\sqrt{130-3^{2}+\frac{96}{2^{4}}}$$

Calculate 3 to the power of 2 and get 9.

$$\sqrt{130-9+\frac{96}{2^{4}}}$$

Subtract 9 from 130 to get 121.

$$\sqrt{121+\frac{96}{2^{4}}}$$

Calculate 2 to the power of 4 and get 16.

$$\sqrt{121+\frac{96}{16}}$$

Divide 96 by 16 to get 6.

$$\sqrt{121+6}$$

Add 121 and 6 to get 127.

$$\sqrt{127}$$