The solution of the equation $2x^2 + 3x - 9 \le 0$ is given by

The complete solution of the inequation $x^2 - 4x < 12$ is

The smallest negative integer satisfying both the quadratic inequalities $x^2 < 4x + 77$ and $x^2 > 4$ is

If $[x]$ denotes the greatest integer not exceeding $x$,then the values of $x$ satisfying $[x]^2-7[x]+12 \leq 0$ are

Assertion $(A)$: $3x^2 - 16x + 4 > -16$ is satisfied for some values of real $x$ in $(0, \frac{10}{3})$.
Reason $(R)$: $ax^2 + bx + c$ and $a$ will have the same sign for some values of $x \in \mathbb{R}$ when $b^2 - 4ac > 0$.
The correct option among the following is

If $x^2-4ax+5+a>0$ for all $x \in R$ whenever $a \in (\alpha, \beta)$,then $4\beta+\alpha=$