Find the quadratic equation, whose solution set is :

$\{3,5\}$

Find the quadratic equation, whose solution set is :

$\{-2,3\}$

Find the value of $\mathrm{x},$ if $\mathrm{a}+1=0$

and ${\mathrm{x}}^2+\mathrm{ax}-6=0.$.

Use the substitution $y = 2x + 3$ to solve for $x$, if $4(2x + 3{)}^2 - (2x + 3) - 14 = 0$.

Without solving the quadratic equation $6x^2-x-2=0,$ find whether $x=\frac{2}{3}$ is a solution of this equation or not.