In order to determine the points of intersection proceed as follow:
Equal both equations:
[tex]4x^2-15x+20=5x-4[/tex]Write the previous equation as an standard quadratic equation:
[tex]\begin{gathered} 4x^2-15x+20=5x-4 \\ 4x^2-15x-5x+20+4=0 \\ 4x^2-20x+24=0 \\ x^2-5x+6=0 \end{gathered}[/tex]to obtain the last equation you divide by 4 both sides.
Now, use the quadratic formula, with a = 1, b = -5 and c = 6, to find the solution for x:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex][tex]\begin{gathered} x=\frac{-(-5)\pm\sqrt[]{(-5)^2-4(1)(6)}}{2(1)} \\ x=\frac{5\pm\sqrt[]{25-24}}{2} \\ x=\frac{5\pm1}{2} \\ x_1=\frac{5-1}{2}=\frac{4}{2}=2 \\ x_2=\frac{5+1}{2}=\frac{6}{2}=3 \end{gathered}[/tex]The previous solutions mean that for the values of x = 2 and x = 3 the given functions intersect each other.
By replacing the values of x into any of the functions, for instance, in
y = 5x - 4, you get:
y = 5(2) - 4 = 10 - 4 = 6
y = 5(3) - 4 = 15 - 4 = 11
Then, the points of intersection are:
(2 , 6)
(3 , 11)
The graph is shown below:
