For what range of k values will the linear-quadratic system below have 2 solutions? y=2x^2+4+5 y=kx-3

So you have
`2x^2+4+5 = kx-3`

`2x^2-kx+12=0` this equation will have two solutions if and
only if its discriminant `D = b^2 - 4ac neq 0` where `a, b, c` are coeffitients of quadratic
equation.

`k^2 - 4 cdot 2 cdot 12 neq 0 implies k neq pm sqrt(96)`

So for `k = pm sqrt(96)` you would have 1 solution and for all other values of `k` you
would have 2 solutions which would be complex if discriminant is negative and real if
discriminant is positive.