Linear-quadratic systems are systems of equations where one equation is quadratic, and the other is linear. And just like with parabolas and the $x$-axis, there can two, one, or no real solutions.
Shockingly, the addition of a quadratic to what you know about systems doesn’t make anything more complicated. Depending on how the system is given to you, either substitute or eliminate through subtraction.
Here’s the first example from the book, which is set up for substitution.
\[\left\{\begin{aligned} &y=3x^2 + 3x -5 \\ &2x-y=3 \end{aligned}\right.\] \[\begin{align*} 2x - (3x^2 + 3x -5) &= 3 \\ -3x^2 - x + 5 &= 3 \\ 3x^2 + x -2 &= 0 \end{align*}\]And you just have another quadratic. You can pick your poison at this point. Factor if you can, and if not complete the square or use the quadratic formula.
For completeness, here’s one where elimination is ideal.
\[\left\{\begin{aligned} y &= x^2 + 9x - 5 \\ y &= 4 - 3x \end{aligned}\right.\]You might want to rewrite it so things are aligned.
\[\left\{\begin{aligned} &y = x^2 &+ 9x - 5 \\ &y = &-3x + 4 \end{aligned}\right.\]And with elimination you end up with
\[0 = x^2 + 12x - 9\]Not factorable, but completing the square would be straightforward since $a=1$.
\[\begin{align*} x^2 + 12x + 36 &= 45 \\ (x+6)^2 &= 45\\ x+6 &=\pm\sqrt{45} \\ x &= -6\pm3\sqrt{5} \end{align*}\]