If -5 is a root of the quadratic equation $2{\mathrm{x}}^2+\mathrm{px}-15=0$ and the quadratic equation $\mathrm{p}\left({\mathrm{x}}^2+\mathrm{x}\right)+\mathrm{k}=0$ has equal roots, then the value of k is

It is given -5 is a root of the equation $2{\mathrm{x}}^2+\mathrm{px}-15=0$ . Therefore, $\mathrm{x}=-5$ satisfies it.
i.e., $2(-5{)}^2-5\mathrm{p}-15=0\Rightarrow 50-5\mathrm{p}-15=0\Rightarrow 5\mathrm{p}=35\Rightarrow \mathrm{p}=7$
Putting $\mathrm{p}=7\text{ in }\mathrm{p}\left({\mathrm{x}}^2+\mathrm{x}\right)+\mathrm{k}=0\text{, we get }7{\mathrm{x}}^2+7\mathrm{x}+\mathrm{k}=0$
The equation will have equal roots, if its discriminant is zero.
i.e, $49-4\times 7\times \mathrm{k}=0\Rightarrow \mathrm{k}=\frac{49}{28}\Rightarrow \mathrm{k}=\frac{7}{4}$.