F4 Amaths Quadratic inequalitiy

Find the range of real values of m for which

(m^2 + 1 )x^2 + 2(m+2)x + 2 > 0 for all real values of x.

since m^2 + 1>0, the curve will open upwards
so that the equation (m^2 + 1 )x^2 + 2(m+2)x + 2 = 0
will have no real roots
∵ discriminant < 0
[2(m+2)]^2 -4 (m^2 + 1 ) (2)< 0
4(m+2)^2 -8 (m^2 + 1 )< 0
4(m^2+4m+4) -8m^2 -8< 0
4m^2+16m+16 -8m^2 -8< 0
-4m^2 +16m+8< 0
-4(m^2 -4m-2)< 0
(m^2 -4m-2)> 0
(m^2 -4m+4-4-2)> 0
(m -2)^2-6> 0
(m -2)^2> 6
(m -2)> √6 or (m -2)< -√6
m > 2+√6 or m < 2-√6

(m^2 + 1 )x^2 + 2(m+2)x + 2 &gt; 0

b^2-4ac&lt;0
so (2(m+2))^2-4(m^2+1)(2)&lt;0
4m^2+8m+8-8m^2-8&lt;0
4m^2-8m&gt;0
m(m-2)&gt;0
m&lt;0 or m&gt;2