Sketch: If you consider this a quadratic in $a$ you'll obtain $a^2-a(1+2x^2)+x^4+x=0,$ which can be factored into $(x^2+x-a)(x^2-x+1-a)=0.$ The interval of when the first factor is real is achieved when the discriminant is greater than or equal to $0,$ in which this case occurs $1+4a \geq 0.$ Solving gives $a \in [-1/4, \infty).$ Similarly, the interval of which the second factor is real is achieved at $4a-3 \geq 0,$ which is achieved when $\boxed{a \in [3/4, \infty)},$ and since the latter is the common interval, it's our answer. Remark. Is there any better way to do this problem other than this, since this seems somewhat tedious.