Explanation
We are given the following system of equations:
[tex]\begin{gathered} y=2x^2-6x+3 \\ y=x-2 \end{gathered}[/tex]We are required to determine the solutions to the given system of equations.
This is achieved thus:
[tex]\begin{gathered} y=2x^2-6x+3 \\ y=x-2 \\ \\ \therefore2x^2-6x+3=x-2 \\ 2x^2-6x-x+3+2=0 \\ 2x^2-7x+5=0 \\ \text{ Using grouping method, we have} \\ 2x^2-2x-5x+5=0 \\ (2x^2-2x)(-5x+5)=0 \\ 2x(x-1)-5(x-1)=0 \\ (x-1)(2x-5)=0 \\ x-1=0;2x-5=0 \\ x=1;2x=5 \\ x_1=1;x_2=\frac{5}{2} \end{gathered}[/tex]Hence, the answer is:
[tex]x_1=1;x_2=\frac{5}{2}[/tex]