+118608 2y^2-12y-17=x
This is a sideways parabola (because it is the y that is squared)
it opens out in the positive direction because the number in front of the y^2 is positive (+2)
When y=0, x=-17
So it must have 2 real roots, if you sketch what I have said so far you will see why.
\(\triangle = b^2-4ac = 144-136=8\)
8 is not a perfect square so there are 2 irrational roots.
The axis of symmetry is y=-b/2a = 12/4 = 3
So the roots, (y intercepts), are,
\(x=\frac{--12\pm\sqrt8}{4}\\ x=\frac{12\pm2\sqrt2}{4}\\ x=\frac{6\pm\sqrt2}{2}\\ \text{So it is a sideways parabola that passes through}\\ (\frac{6+\sqrt2}{2},0),\;\;\;(\frac{6-\sqrt2}{2},0),\;\;\;and \;\;\;(-17,0)\)
At the vertex y=3 sub that in to get the value of x
