[tex]\bf Vom\ nota\ \sqrt{x+1}=t,\ t>0,\ x+1=t^2 \Rightarrow 2x+3=2t^2+1\\ \\ Acum,\ ecua\c{\bf t}ia\ se\ scrie:\\ \\ \sqrt{2t^2+1}=5-t \Rightarrow 2t^2+1=25-10t+t^2 \Rightarrow t^2+10t-24=0 \Rightarrow \\ \\ \Rightarrow t^2+10t+25-49=0 \Rightarrow (t+5)^2 -7^2=0 \Rightarrow (t+5-7)(t+5+7)=0 \Rightarrow \\ \\ \Rightarrow (t-2)(t+12)=0 \Rightarrow t=2,\ solu\c{\bf t}ie\ acceptabil\breve a,\ deoarece\ t>0\\ \\ t=2 \Rightarrow \sqrt{x+1}=2 \Rightarrow x+1=4 \Rightarrow x=3\\ \\ Verificarea\ confirm\breve a.[/tex]
