[tex]f(x)=x+3 \\ \\ f(x)=x^2 \\ \\ \Rightarrow x+3=x^2. \\ \\ x^2-x-3=0 \\ \\ \Delta=b^2-4ac=(-1)^2-4 \cdot 1 \cdot (-3)=1+12=13. \\ \\ x_1= \frac{-b + \sqrt{\Delta}}{2a}= \frac{1+\sqrt{13}}{2}. \\ \\ x_2= \frac{-b- \sqrt{\Delta}}{2a} = \frac{1- \sqrt{13}}{2}. [/tex]
[tex]\underline{Solutie}:~x \in \Big \{ \frac{1+ \sqrt{13}}{2}~;~ \frac{1- \sqrt{13}}{2} \Big \}. [/tex]
