Solve sqrt{left(3-2right)^2+left(-216619-1right)^2+left(10-2right)^2}

$$\sqrt{ { \left(3-2 \right) }^{ 2 } + { \left(-216619-1 \right) }^{ 2 } + { \left(10-2 \right) }^{ 2 } }$$

Evaluate

$$\sqrt{46924224465}\approx 216620.000150032$$

Solution Steps

Subtract 2 from 3 to get 1.

$$\sqrt{1^{2}+\left(-216619-1\right)^{2}+\left(10-2\right)^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$\sqrt{1+\left(-216619-1\right)^{2}+\left(10-2\right)^{2}}$$

Subtract 1 from -216619 to get -216620.

$$\sqrt{1+\left(-216620\right)^{2}+\left(10-2\right)^{2}}$$

Calculate -216620 to the power of 2 and get 46924224400.

$$\sqrt{1+46924224400+\left(10-2\right)^{2}}$$

Add 1 and 46924224400 to get 46924224401.

$$\sqrt{46924224401+\left(10-2\right)^{2}}$$

Subtract 2 from 10 to get 8.

$$\sqrt{46924224401+8^{2}}$$

Calculate 8 to the power of 2 and get 64.

$$\sqrt{46924224401+64}$$

Add 46924224401 and 64 to get 46924224465.

$$\sqrt{46924224465}$$