Quadratic in a quadratic?

According to the question the discriminant $b^2-4ac = 0$ , because it has equal roots.(b=m) and $c=m^2+3m-32$ . $\therefore b^2-4ac = m^2+4(m^2+3m-32) = 0 \\\implies 5m^2+12m-128 = 0\\ \\m=\dfrac{-12\pm\sqrt{144+2560}}{2\times5}\\=\dfrac{-12\pm 52}{10}$

Hence $\boxed{m=4}$ or $m=-6\dfrac 25$ , of which 4 is positive.

If $x^2 + mx - (m^2 + 3m - 32)$ has two equal roots $\Rightarrow x^2 + mx - (m^2 + 3m - 32) = (x - a)^2$ for some $a \in \mathbb{C}\Rightarrow \text{m = - 2a and} -(m^2 + 3m - 32) = a^2\Rightarrow$ $-5a^2 + 6a + 32 = 0 \Rightarrow \text{(m positive integer)} a = -2 \Rightarrow m =4$

Expressing the equation in the form $a{x}^{2}+bx+c=0$ , we get that ${ x }^{ 2 }+mx+(-{ m }^{ 2 }-3m+32)=0$ and if the discriminant is ${ b }^{ 2 }-4ac=0$ , then the roots are equal. Therefore, we will substitute the values of $a$ , $b$ , and $c$ . $\begin{aligned} {b}^{2}-4ac=0 \\{(m})^{2}-4(1)(-{m}^{2}-3m+32)=0 \\{m}^{2}+4{m}^{2}+12m-128=0 \\5{m}^{2}+12m-128=0 \\ (5m+32)(m-4)=0 \end{aligned}$

Hence, $m=\{ -\frac { 32 }{ 5 } ,4\}$ but the question asks for the positive integral value of $m$ so $\boxed { m=4 }$ .

nice solution Chebrolu.