Since f(x) = y and g(x) = y, you can set them equal to each other:
[tex]\begin{gathered}\sf \: x²\:+\:3x\:-\:2 = 2x\:+\:3 \\ \\ \sf \: x²\:+\:x\:-\:5 = 0 \end{gathered}[/tex]
Solving the quadratic equation with the quadratic formula:
[tex]\begin{gathered}\sf \: x = \dfrac{-1\:±\:\sqrt{1²\:-\:4\:·\:1\:·\:(-5)}}{2\:·\:1} \\ \\ \sf \: = \dfrac{-1\:±\:\sqrt{1\:+\:20}}{2} \\ \\ \sf \: = \dfrac{-1\:±\:21}{2}, \end{gathered}[/tex]
so
[tex]\begin{gathered}\sf \: x₁ = \dfrac{-1\:-\:\sqrt{21}}{2} ≈ -2.79, \\ \\ \sf \: x₂ = \dfrac{-1\:±\:\sqrt{21}}{2} ≈ 1.79. \end{gathered}[/tex]
[tex]\:[/tex]
Insert x₁ and x₂ to g(x) = 2x + 3 because it's the simplest expression. You can also insert to f(x).
[tex]\sf{y₁\:=\:g(-2.79)\:=\:2\:·\:(-2.79)\:+\:3\:=\:-2.58}[/tex]
[tex]\sf{y₂\:=\:g(1.79)\:=\:2\:·\:1.79\:+\:3\:=\:6.58}[/tex]
[tex]\:[/tex]
The intersections between f(x) and g(x) is
[tex]\begin{gathered}\sf \: (x₁, y₁) = (-2.79, -2.58), \\ \\ \sf \: (x₂, y₂) = (1.79, 6.58). \\ \\ \sf \: = \dfrac{-5\:±\:1}{2} \end{gathered}[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:\:\:\boxed{\sf{See\:the\:picture\:for\:the\:graph}}[/tex]
[tex]\:[/tex]
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