You have to equal the two Ys, meaning their values as well or you can find the value of the first x and then plug it in the second equation. There are many ways to solve this.
y=x+3 and y=2x^2
y=y=>x+3=2x^2=>2x^2-x-3=0
You can use any tools you know to solve this quadratic equation but as for me, I will use Delta
Delta=b^2-4ac, with a=2, b=-1 and c=-3
Delta=(-1)^2-4(2)(-3)=25=> sqrt Delta=+-5
x_1=(-b+sqrt Delta)/(2a) and x_2=(-b-sqrt Delta)/(2a)
x_1=(1+5)/(4)=6/4=3/2 and x_2=(1-5)/(4)=-1
x_1=3/2 and x_2=-1
To find y, all you have to do is to plug the x values in either of the two equations. I will plug in both just to show you that it does not matter which one you chose.
With the first equation y=x+3
For x=3/2=>y=3/2+3=(3+6)/2=9/2
For x=-1=>y=-1+3=2
With the second equation y=2x^2
For x=3/2=>y=2(3/2)^2=1 color (red) cancel 2(9/(2 color (red) cancel4))=9/2
For x=-1=>y=2(-1)^2=2
Therefore, your solution is (3/2,9/2) and (-1,2)
Hope this helps :)