The given system is
[tex]\mleft\{\begin{aligned}y-2x^2=4x+1 \\ y=14x-7\end{aligned}\mright.[/tex]Let's replace the second equation with the first one.
[tex]14x-7-2x^2=4x+1[/tex]Now, we solve the quadratic equation.
[tex]\begin{gathered} 14x-7-2x^2-4x-1=0 \\ -2x^2+10x-8=0 \\ -2(x^2-5x+4)=0 \\ x^2-5x+4=0 \end{gathered}[/tex]Then, we have to find two numbers whose product is 4 and whose sum is 5. Those numbers are 4 and 1.
[tex]\begin{gathered} (x-4)(x-1)=0 \\ x_1=4 \\ x_2=1 \end{gathered}[/tex]We use these values to find y.
[tex]\begin{gathered} y_1=14\cdot4-7=49 \\ y_2=14\cdot1-7=7 \end{gathered}[/tex]