Solve D=sqrt{3^2+2^2+2,8^2}

$$D = \sqrt { 3 ^ { 2 } + 2 ^ { 2 } + 2,8 ^ { 2 } }$$

Solve for D

$$D=\sqrt{141}\approx 11.874342087$$

Steps for Solving Linear Equation

Calculate 3 to the power of 2 and get 9.

$$D=\sqrt{9+2^{2}+2\times 8^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$D=\sqrt{9+4+2\times 8^{2}}$$

Add 9 and 4 to get 13.

$$D=\sqrt{13+2\times 8^{2}}$$

Calculate 8 to the power of 2 and get 64.

$$D=\sqrt{13+2\times 64}$$

Multiply 2 and 64 to get 128.

$$D=\sqrt{13+128}$$

Add 13 and 128 to get 141.

$$D=\sqrt{141}$$

Assign D

$$D≔\sqrt{141}$$