Question 29299
Try this:
{{{y = 2x^2}}} and {{{y = x+3}}} Since y = y you can write:
{{{2x^2 = x+3}}} Simplify.
{{{2x^2-x-3 = 0}}} Solve this quadratic by factoring.
{{{(2x-3)(x+1) = 0}}} Apply the zero products principle.
{{{2x-3 = 0}}} and/or {{{x+1 = 0}}}
If {{{2x-3 = 0}}} then {{{2x = 3}}} and {{{x = 3/2}}}
If {{{x+1 = 0}}} then {{{x = -1}}}

We have found the x-coordinates of the solutions. To find the y-coordinates, substitute the x-values into either of the original equations and solve for y.

{{{x = 3/2}}} Substitute into {{{y = x+3}}}
{{{y = 3/2 + 3}}}
{{{y = 3/2 + 6/2}}}
{{{y = 9/2}}}

{{{x = -1}}} Substitute into {{{y = 2x^2}}}
{{{y = 2(-1)^2}}}
{{{y = 2}}}

The solutions are:
(3/2, 9/2) and (-1, 2)

Why are there two solutions? Because you have a parabola {{{y = 2x^2}}} transversed by a line {{{y = x+3}}}

Here's a graph showing the two intersections/solutions.
{{{graph(300,200,-5,5,-5,5,2x^2,x+3)}}}