Notice how easily we can isolate #x# in the equation #y^2=x+3#. We can subtract #3# from both sides of the equation to reveal that #color(blue)(x=y^2-3)#. With this, we can replace #color(blue)x# in the second equation, #color(blue)x-2y=12#, with the value that we know #color(blue)x# is equal to: #color(blue)(y^2-3)#.

This gives us: #" "color(blue)x-2y=12" "=>" "(color(blue)(y^2-3))-2y=12#

Continuing to solve the resulting equation, we obtain #y^2-2y-15=0#, which can be factored into #(y-5)(y+3)=0# and solved giving **two** solutions for #y#, which are that #color(red)(y=5)# and #color(green)(y=-3)#.

Each of the values of #y# can be plugged into *either* one of the two equations to find their corresponding values of #x#. I will choose the first equation, although you will receive the same results should you choose to input your values of #y# into the second.

Solving for #x# when #color(red)(y=5)#, we obtain:

#color(red)y^2=x+3" "=>" "color(red)5^2=x+3" "=>" "color(red)(x=22)#

Since #color(red)(x=22)# and #color(red)(y=5)#, we have a solution point at #(color(red)22,color(red)5)#.

And for when #color(green)(y=-3)#, we obtain:

#color(green)y^2=x+3" "=>" "(color(green)(-3))^2=x+3" "=>" "color(green)(x=6)#

Thus there is also a solution point at #(color(green)6,color(green)(-3))#.

Graphing the two equations should reveal a sideways parabola with a line intersecting it at the points #(22,5)# and #(6,-3)#:

graph{(y^2-x-3)(x-2y-12)=0 [-5.14, 30.9, -7.66, 10.36]}