Solve 7*3+[6+2*(2^3div4+3*2)-7*sqrt{4}]+9=3

$$7 \cdot 3 + [ 6 + 2 * ( 2 ^ { 3 } \div 4 + 3 * 2 ) - 7 * \sqrt { 4 } ] + 9 = 3$$

Verify

Solution Steps

Multiply both sides of the equation by 4.

$$28\times 3+24+8\left(\frac{2^{3}}{4}+3\times 2\right)-28\sqrt{4}+36=12$$

Multiply 28 and 3 to get 84.

$$84+24+8\left(\frac{2^{3}}{4}+3\times 2\right)-28\sqrt{4}+36=12$$

Add 84 and 24 to get 108.

$$108+8\left(\frac{2^{3}}{4}+3\times 2\right)-28\sqrt{4}+36=12$$

Calculate 2 to the power of 3 and get 8.

$$108+8\left(\frac{8}{4}+3\times 2\right)-28\sqrt{4}+36=12$$

Divide 8 by 4 to get 2.

$$108+8\left(2+3\times 2\right)-28\sqrt{4}+36=12$$

Multiply 3 and 2 to get 6.

$$108+8\left(2+6\right)-28\sqrt{4}+36=12$$

Add 2 and 6 to get 8.

$$108+8\times 8-28\sqrt{4}+36=12$$

Multiply 8 and 8 to get 64.

$$108+64-28\sqrt{4}+36=12$$

Add 108 and 64 to get 172.

$$172-28\sqrt{4}+36=12$$

Calculate the square root of 4 and get 2.

$$172-28\times 2+36=12$$

Multiply -28 and 2 to get -56.

$$172-56+36=12$$

Subtract 56 from 172 to get 116.

$$116+36=12$$

Add 116 and 36 to get 152.

$$152=12$$

Compare 152 and 12.

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