If the inequality `2ax^2 + (a+3)x + a -1 > 0` is to have no solution
this implies that the quadratic `2ax^2 + (a+3)x + a - 1` is strictly negative
ie that the quadratic `-2ax^2 - (a+3)x - a + 1` is strictly positive.
A quadratic is strictly positive when it has no real roots, that is, when it's determinant is negative.
The determinant is given by `B^2 - 4AC` where `A ` is the coefficient of `x^2`, `B` is the coefficient of `x` and `C` is the zeroth order term.
Here, `A = -2a`, `B = -(a +3)` and `C = 1 -a`
Therefore the determinant is given by `B^2 - 4AC = (a+3)^2 - 4(-2a)(1-a)`
This is negative if `a^2 + 6a + 9 - 4(-2a + 2a^2) = -7a^2 + 14a + 9 < 0`
The roots of `-7a^2 + 14a + 9 = 0` are given by (using the quadratic formula)
`(-14 +- sqrt(14^2 - 4(-7)(9)))/(-14)`
`= 1 +- sqrt(448)/14 = 1 +- sqrt(32/14) = 1 +- sqrt(16/7) = 1 +- 4/sqrt(7)`
Since the determinant is a negative quadratic it is concave and so is negative below its first root and above its second root.
Therefore, the inequality has no real solutions when
`a: a in (-oo,1-4/sqrt(7))uu(1+4/sqrt(7),oo)`