Solve d=(2)^2+(-3)^2+(m-3)^2

$$d = ( 2 ) ^ { 2 } + ( - 3 ) ^ { 2 } + ( m - 3 ) ^ { 2 }$$

Solve for d

$$d=\left(m-3\right)^{2}+13$$

Solution Steps

Calculate 2 to the power of 2 and get 4.

$$d=4+\left(-3\right)^{2}+\left(m-3\right)^{2}$$

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

$$d=4+9+\left(m-3\right)^{2}$$

Add 4 and 9 to get 13.

$$d=13+\left(m-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(m-3\right)^{2}.

$$d=13+m^{2}-6m+9$$

Add 13 and 9 to get 22.

$$d=22+m^{2}-6m$$

Solve for m (complex solution)

$$m=-\sqrt{d-13}+3$$

$$m=\sqrt{d-13}+3$$

Solve for m

$$m=-\sqrt{d-13}+3$$

$$m=\sqrt{d-13}+3\text{, }d\geq 13$$