First, we use the given functions to solve the equation f(x) = g(x)
[tex]x^2+13=3x+14[/tex]
Now, we move all the terms to the left side
[tex]\begin{gathered} x^2-3x+13-14=0 \\ x^2-3x-1=0 \end{gathered}[/tex]
Then, we use the quadratic formula
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Where a = 1, b = -3, and c = -1.
[tex]\begin{gathered} x=\frac{-(-3)\pm\sqrt[]{(-3)^2-4\cdot(1)(-1)}}{2\cdot1} \\ x=\frac{3\pm\sqrt[]{9+4}}{2}=\frac{3\pm\sqrt[]{13}}{2} \\ x_1=\frac{3+\sqrt[]{13}}{2}\approx3.3 \\ x_2=\frac{3-\sqrt[]{13}}{2}\approx-0.3 \end{gathered}[/tex]
Hence, the positive solution is 3.3
