Đặt $A=\left(x-3\right)^2+\left(x+4\right)^2$

$=x^2-6x+9+x^2+8x+16$

$=2x^2+2x+25$

$=2\left(x^2+x+\dfrac{25}{2}\right)$

$=2\left(x^2+x+\dfrac{1}{4}+\dfrac{49}{4}\right)$

$=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{49}{2}$

Ta có: $2\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Leftrightarrow2\left(x+\dfrac{1}{2}\right)^2+\dfrac{49}{2}\ge\dfrac{49}{2}\forall x$

Dấu "=" xảy ra khi $x+\dfrac{1}{2}=0$ hay $x=-\dfrac{1}{2}$

Vậy AMIN = $\dfrac{49}{2}$ khi x = $-\dfrac{1}{2}$.