Solve {l}{1^2-(sqrt{3})^2}{6.01-(sqrt{3})^2}{6.01-3}

$$\left. \begin{array} { l } { 1 ^ { 2 } - ( \sqrt { 3 } ) ^ { 2 } } \\ { 6.01 - ( \sqrt { 3 } ) ^ { 2 } } \\ { 6.01 - 3 } \end{array} \right.$$

Sort

$$-2,\ 3.01,\ 3.01$$

Solution Steps

Calculate 1 to the power of 2 and get 1.

$$sort(1-\left(\sqrt{3}\right)^{2},6.01-\left(\sqrt{3}\right)^{2},6.01-3)$$

The square of \sqrt{3} is 3.

$$sort(1-3,6.01-\left(\sqrt{3}\right)^{2},6.01-3)$$

Subtract 3 from 1 to get -2.

$$sort(-2,6.01-\left(\sqrt{3}\right)^{2},6.01-3)$$

The square of \sqrt{3} is 3.

$$sort(-2,6.01-3,6.01-3)$$

Subtract 3 from 6.01 to get 3.01.

$$sort(-2,3.01,6.01-3)$$

Subtract 3 from 6.01 to get 3.01.

$$sort(-2,3.01,3.01)$$

The list values are already in order.

$$-2,3.01,3.01$$

Evaluate

$$-2,\ 3.01,\ 3.01$$