\(\dfrac{A}{3}=x^2+\dfrac{4}{3}x-\dfrac{5}{3}\)
\(\dfrac{A}{3}=x^2+\dfrac{4}{3}x+\dfrac{4}{9}-\dfrac{19}{9}\)
\(\dfrac{A}{3}=\left(x+\dfrac{2}{3}\right)^2-\dfrac{19}{9}\ge-\dfrac{19}{9}\)
\(\dfrac{A}{3}\ge-\dfrac{19}{9}\Leftrightarrow A\ge-\dfrac{19}{3}\)
\("="\Leftrightarrow x=-\dfrac{2}{3}\)