We are given `y=2x^2+4x+5` The slope of the tangent line is given by `y'=4x+4` .
The system will have two solutions if the line intersects the parabola twice. It will do so between the values where the line is tangent to the parabola.
A point on the parabola has coordinates `(x,2x^2+4x+5)` . The slope between this point and the point (0,-3) (Which lies on any line y=kx-3) is:
`m=(2x^2+4x+5-(-3))/(x-0)=(2x^2+4x+8)/x` . If the linewith this slope is tangent to the curve its slope is also give by m=4x+4.
If x=2 then the point on the parabola is (2,21) and the slope between (2,21) and (0,-3) is 12, while the slope of the tangent line at x=2 is 4(2)+4=12
Also, if x=-2 the pointon the parabola is (-2,5) and the slope from (-2,5) to (0,-3) is -4, while the slope of the tangent line at x=-2 is 4(-2)+4=-4.
If x=2, then k=12 and if x=-2 then k=-4. So any value of k such that -4<k<12 will cause the line y=kx-3 to intersect the parabola twice.
The graphs of the parabola and the two tangent lines: