[tex]\it Notez\ \sqrt{\mathcal{A}_b}=x,\ iar\ ecua\c{\it t}ia\ devine:\\ \\ 1110=\dfrac{10}{3}(x^2+144+12x)|_{\cdot\frac{3}{10}}\Rightarrow 333=x^2+144+12x}|_{-144} \Rightarrow\\ \\ \Rightarrow 189=x^2+12x \Rightarrow x^2+12x=189 \Rightarrow x(x+12) = 9\cdot21 \Rightarrow\\ \\\Rightarrow x(x+12) = 9\cdot(9+12)\Rightarrow x = 9 \Rightarrow \sqrt{\mathcal{A}_b}=9 \Rightarrow \mathcal{A}_b=81.[/tex]
