How do you solve this ?

x^2+8y=16
 3x+4y=12

Multiply the second equation by 2  .....this gives us     6x + 8y  = 24

Rearrange this to   ....    8y = 24 - 6x      and substitute this into the first equation....so we have

x^2 + [24 - 6x] = 16     simplify

x^2 - 6x + 8  = 0       factor

(x - 4) (x -2) = 0     and setting both factors to 0, we have that x = 2 and x = 4

And using the second equation to find y in both cases

3(2) + 4y = 12              3(4) + 4y  = 12

6 + 4y = 12                   12 + 4y   = 12

4y = 6                             4y  =  0

y = 3/2                             y = 0

So....the intersection points are (2, 3/2)  and ( 4, 0 )

Here's a graph  :  https://www.desmos.com/calculator/ac9ayppsom