At the point of intersection of \(y=2 x+1\) and \(y=2 x^2+5 x-1\), we can write,
\begin{array}{l}
2 x+1=2 x^2+5 x-1 \\
2 x^2+5 x-2 x-1-1=0 \\
2 x^2+3 x-2=0 \\
2 x^2+4 x-x-2=0 \\
2 x(x+2)-1(x+2)=0 \\
(2 x-1)(x+2)=0 \\
\text { Either } 2 x-2=0 \text { or } x+2=0 \\
2 x=1 \text { or } x=-2, x=1 / 2 \text { or } x=-2
\end{array}
To get the value of \(y\) for each value of \(x\), we substitute for \(x\) in either of the two equations; note that using the equation of the line will be easier when \(x=1 / 2, y=2(1 / 2)+1=1+1=2\)
when \(x=-2, y=2(-2)+1=-4+1=-3\)
then, the points of intersection are
\((1 / 2,2)\) and \((-2,-3)\)