Solve sqrt{11}+2^{2}*3-5*(5^{2}-2^{3}*3+1)^2]^4:(6^2-3^3)+

$$\sqrt { 11 } + 2 ^ { 2 } \times 3 - 5 \times ( 5 ^ { 2 } - 2 ^ { 3 } \times 3 + 1 ) ^ { 2 } ] ^ { 4 } : ( 6 ^ { 2 } - 3 ^ { 3 } ) +$$

Evaluate

$$\sqrt{11}-8\approx -4.68337521$$

Solution Steps

Calculate 2 to the power of 2 and get 4.

$$\sqrt{11}+4\times 3-5\left(5^{2}-2^{3}\times 3+1\right)^{2}$$

Multiply 4 and 3 to get 12.

$$\sqrt{11}+12-5\left(5^{2}-2^{3}\times 3+1\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$\sqrt{11}+12-5\left(25-2^{3}\times 3+1\right)^{2}$$

Calculate 2 to the power of 3 and get 8.

$$\sqrt{11}+12-5\left(25-8\times 3+1\right)^{2}$$

Multiply 8 and 3 to get 24.

$$\sqrt{11}+12-5\left(25-24+1\right)^{2}$$

Subtract 24 from 25 to get 1.

$$\sqrt{11}+12-5\left(1+1\right)^{2}$$

Add 1 and 1 to get 2.

$$\sqrt{11}+12-5\times 2^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\sqrt{11}+12-5\times 4$$

Multiply 5 and 4 to get 20.

$$\sqrt{11}+12-20$$

Subtract 20 from 12 to get -8.

$$\sqrt{11}-8$$