You shouldn't need The Wizard for this one

Let $I(\alpha)=\int_{-\infty}^{\infty} (-a^2x^2+bx+c)e^{\alpha(-a^2x^2+bx+c)}dx.$

Then $\int I(\alpha)d\alpha = \int_{-\infty}^{\infty}e^{\alpha(-a^2x^2+bx+c)}dx+C$

$=e^{\alpha\Big(c+\frac{b^2}{4a^2}\Big)}\int_{-\infty}^{\infty} e^{-\alpha a^2\Big(x-\frac{b}{2a^2}\Big)^2}dx+C$

$=e^{\alpha\Big(c+\frac{b^2}{4a^2}\Big)}\sqrt{\frac{\pi}{\alpha a^2}}+C$

$I(\alpha)=\frac{d}{d\alpha}\int I(\alpha)d\alpha=\Big(c+\frac{b^2}{4a^2}\Big)e^{\alpha\Big(c+\frac{b^2}{4a^2}\Big)}\sqrt{\frac{\pi}{\alpha a^2}}+e^{\alpha\Big(c+\frac{b^2}{4a^2}\Big)}\Big(-\frac{1}{2}\sqrt{\frac{\pi}{\alpha^3a^2}}\Big)$

$=\frac{1}{\sqrt{\alpha}}e^{\alpha\Big(c+\frac{b^2}{4a^2}\Big)}\frac{\sqrt{\pi}}{a}\Big[c+\frac{b^2}{4a^2}-\frac{1}{2\alpha}\Big].$

Letting $\alpha=25$ leads to the result.

A cheeky solution (lol)

Since we know that the equation above holds for all constants $a,b,c$ , we can pick $a,b,c$ which make our life easier, let's pick $c=b=0$ and $a=1$ !

The desired integral reduces to $\int_{-\infty}^{\infty} -x^2e^{-25x^2} \, dx=\cancelto{=0}{\left. -\frac{x}{50} e^{-25x^2} \right|_{-\infty}^{\infty}}-\frac{1}{50}\int_{-\infty}^{\infty} e^{-25x^2} \, dx=-\frac{1}{50\cdot 5} \int_{-\infty}^{\infty} e^{-u^2} \, du=-\frac{\sqrt \pi}{50 \cdot 5}$ with a substitution $u=5x$ at the end and based on the given equation, our result should be $\displaystyle -\frac{\sqrt{\pi}}{m \cdot n}$ and we can should take $m+n=5+50=55$ .