The values of a for  which the  inequality $x^2+|x-a|-3<0$ is satisfied by at least one negative x

Hence when $a_1<a<a_2,$ at least one of the arms of  the mod –graph will be below the parabola for some negative values of x .we  can using coordation  geometry ,easily evaluate $a_{1}and{a}_2$ to be $-\frac{13}{4}$  and 3 respectively . This is left to the reader as an exercise. Therefore ,$a\in (-\frac{13}{4},3)$