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

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

Resolver d

$$d=\left(m-3\right)^{2}+1$$

Pasos da solución

Calcula 2 á potencia de 2 e obtén 4.

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

Resta 3 de 4 para obter 1.

$$d=1+\left(m-3\right)^{2}$$

Usar teorema binomial \left(a-b\right)^{2}=a^{2}-2ab+b^{2} para expandir \left(m-3\right)^{2}.

$$d=1+m^{2}-6m+9$$

Suma 1 e 9 para obter 10.

$$d=10+m^{2}-6m$$

Resolver m

$$m=-\sqrt{d-1}+3$$

$$m=\sqrt{d-1}+3\text{, }d\geq 1$$