jmc

Factoring our original equation: $$3x^2+10x-25=(3x-5)(x+5)=0$$Therefore, the two solutions are $3x-5=0$ and $x+5=0$. $$\begin{array}{rl}3x-5& =0\\ 3x& =5\\ x& =\frac{5}{3}\end{array}$$ $$\begin{array}{rl}x+5& =0\\ x& =-5\end{array}$$Therefore, either $d$ or $e$ could equal either $\frac{5}{3}$ or $-5$. However, $(d-e{)}^2=(e-d{)}^2$ as shown below. $$\begin{array}{rl}(d-e{)}^2& =\{(-1)(e-d){\}}^2\\ & =(-1{)}^2(e-d{)}^2\\ & =(e-d{)}^2\end{array}$$Setting $\frac{5}{3}=d$ and $-5=e$: $$\begin{array}{rl}(d-e{)}^2& ={\left(\frac{5}{3}-(-5)\right)}^2\\ & ={\left(\frac{5}{3}+5\right)}^2\\ & ={\left(\frac{20}{3}\right)}^2\\ & =\boxed{\frac{400}{9}}\end{array}$$