Answer:
graph?
Step-by-step explanation:
![y=2x^2+8x-9\\D=b^2-4ac\\D=64-4(2)(-9)\\D=64+72 > 0](https://tex.z-dn.net/?f=y%3D2x%5E2%2B8x-9%5C%5CD%3Db%5E2-4ac%5C%5CD%3D64-4%282%29%28-9%29%5C%5CD%3D64%2B72%20%3E%200)
There are 2 roots so the only way to complete the square is,
![y=2x^2+8x-9\\y=2[(x^2+4x)]-9\\y=2[(x^2+4x+4)-4]-9\\y=2[(x+2)^2-4]-9\\y=2(x+2)^2-8-9\\y=2(x+2)^2-17](https://tex.z-dn.net/?f=y%3D2x%5E2%2B8x-9%5C%5Cy%3D2%5B%28x%5E2%2B4x%29%5D-9%5C%5Cy%3D2%5B%28x%5E2%2B4x%2B4%29-4%5D-9%5C%5Cy%3D2%5B%28x%2B2%29%5E2-4%5D-9%5C%5Cy%3D2%28x%2B2%29%5E2-8-9%5C%5Cy%3D2%28x%2B2%29%5E2-17)
Just factor 2 out of 2x^2+8x (just ignore the -9) then find the number that will make the terms be able to complete the square.
then complete the square and multiply 2 inside the brackets.
subtraction as you already get the vertex form and know how to complete the square.
Vertex Form: ![y=2(x+2)^2-17](https://tex.z-dn.net/?f=y%3D2%28x%2B2%29%5E2-17)
3x = 5(x-4)
I am not sure about the equal sign though.