We have the equation:
[tex]\begin{gathered} -x^2+2x+4=1 \\ -x^2+2x+4-1=0 \\ -x^2+2x+3=0 \end{gathered}[/tex]This is a quadratic equation, so we can solve it like this:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ a=-1,b=2,c=3 \\ \\ x_{1,2}=\frac{-2\pm\sqrt{2^2-4\cdot(-1)\cdot3}}{2\cdot(-1)}=\frac{-2\pm\sqrt{4+12}}{-2}=\frac{-2\pm\sqrt{16}}{-2}=\frac{-2\pm4}{-2} \\ \\ x_1=\frac{-2+4}{-2}=\frac{2}{-2}=-1 \\ \\ x_2=\frac{-2-4}{-2}=\frac{-6}{-2}=3 \end{gathered}[/tex]The solutions to this equation are x=-1 and x=3.
So it is TRUE that 3 is a solution to the equation.