It is all about quadratic equations

$\begin{aligned} f(x)&=2x^2+px-15 \\ g(x)&=p(x^2+x)+k=px^2+px+k\end{aligned}$ Since $x=-5$ is a root of the quadratic $f(x)$ , we have, $\begin{aligned} f(-5)&=0 \\ 2(-5)^2+p(-5)-15&=0 \\ \therefore \; p&=7 \end{aligned}$ Also, the quadratic $g(x)=px^2+px+k=7x^2+7x+k$ has equal roots, implying that its discriminant must be zero. $\begin{aligned} \Delta& =0 \\ 7^2-4\cdot 7\cdot k&=0 \\ 49-28k&=0 \end{aligned}$ $\therefore \; k=\frac{49}{28}=\boxed{\frac{7}{4}}$

Since $x=-5$ is a root to $2x^2 + qx - 15 = 0$ , $\implies 2 (-5)^2 - 5p - 15 = 0 \implies 10 - p - 3 = 0 \implies p = 7$ . Then we have:

$\begin{aligned} p(x^2 + x) + k & = 0 \\ 7(x^2 + x) + k & = 0 \\ x^2 + x + \frac k7 & = 0 \\ \left(x + \frac 12\right)^2 - \frac 14 + \frac k7 & = 0 \end{aligned}$

For $p(x^2+x)+k = 0$ to have equal roots of $x = - \dfrac 12$ , $\implies \dfrac k7 = \dfrac 14 \implies k = \boxed{\frac 74}$ .

Hint : 2(-5)2 + p x ( -5) - 15 = 0 implies p = 7

Now, p(x2 + x ) + k = 0 ,, substitute the value of p in the equation

Apply, D = B2 - 4ac

Solution : p = 7 , k = 7/4