Solve the given two equations and choose the correct answer from the given options.
$I.$ $16x^2 + 20x + 6 = 0$
$II.$ $10y^2 + 38y + 24 = 0$

If $a < 0$,then the inequality $ax^2 - 2x + 4 > 0$ has the solution represented by:

The condition for the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$ to be equal is

Find the value of $k$ so that the sum of the roots of the equation $3 x^{2}+(2 k+1) x-k-5=0$ is equal to the product of the roots:

If $\alpha, \beta$ and $\gamma$ are three consecutive terms of a non-constant $G.P.$ such that the equations $\alpha x^2 + 2\beta x + \gamma = 0$ and $x^2 + x - 1 = 0$ have a common root,then $\alpha(\beta + \gamma)$ is equal to

Let $P(x) = x^3 - ax^2 + bx + c$ where $a, b, c \in R$ has integral roots such that $P(6) = 3$,then '$a$' cannot be equal to