Solve d=sqrt{(-3-0)^2+(3-6)^2}

$$d = \sqrt { ( - 3 - 0 ) ^ { 2 } + ( 3 - 6 ) ^ { 2 } }$$

Solve for d

$$d=3\sqrt{2}\approx 4.242640687$$

Steps for Solving Linear Equation

Subtract 0 from -3 to get -3.

$$d=\sqrt{\left(-3\right)^{2}+\left(3-6\right)^{2}}$$

Calculate -3 to the power of 2 and get 9.

$$d=\sqrt{9+\left(3-6\right)^{2}}$$

Subtract 6 from 3 to get -3.

$$d=\sqrt{9+\left(-3\right)^{2}}$$

Calculate -3 to the power of 2 and get 9.

$$d=\sqrt{9+9}$$

Add 9 and 9 to get 18.

$$d=\sqrt{18}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$d=3\sqrt{2}$$

Assign d

$$d≔3\sqrt{2}$$