A simple problem

I kinda just calculated the discriminant. Discriminant<0, so (after some calculation) (m+6)(m-4)>0. m>5

$(m-2)x^{ 2 }+8x+(m+4)>0$ . We see by plugging in that m=1 and m=2 do not make this statement true for all real x, thus m-2 must be greater than 0 for this inequality to be true for all x. Therefore, we can divide by m-2 without having to flip the inequality. $x^{ 2 }+\frac { 8 }{ m-2 } x+\frac { m+4 }{ m-2 } >0$ . In other words, the parabola $P(x)=x^{ 2 }+\frac { 8 }{ m-2 } x+\frac { m+4 }{ m-2 }$ is positive everywhere. All that must be true for that to be the case, is that it’s vertex is above the x-axis, because the vertex is the minimum of a parabola with a positive leading coefficient. $P(x)=(x+\frac { 4 }{ m-2 } )^{ 2 }+\frac { m^{ 2 }+2m-24 }{ (m-2)^{ 2 } }$ . In order for the vertex of the parabola to be above the x-axis we need it’s y-value to be greater than 0. $\frac { m^{ 2 }+2m-24 }{ (m-2)^{ 2 } } >0$ or $(m+6)(m-4)>0$ . Using a sign table we find that this is true for the intervals $(-\infty ,-6)\cup (4,\infty )$ . We recall that m is greater than 2 and a natural number, so the lowest value of m which works is 5 (note that 4 was exclusive).