We have $2\sqrt{t}=5\sqrt{s}-10$. Square both sides.
We get $4t=(5\sqrt{s}-10)^2$. But $4t=2s+7$.
So we are at $2s+7=(5\sqrt{s}-10)^2$.
Let $w=\sqrt{s}$. Then $2w^2+7=5^2(w-2)^2=25w^2-100w+100$. Now we can use the Quadratic Formula, though in fact the polynomial $23w^2-100w+93$ factors pleasantly, and we get $w=3$ or $w=31/23$, giving $s=9$ or $s=(31/23)^2$. Now we can find the accompanying $t$.
Remember that we squared during the solving process. This may have resulted in "extraneous" roots that are not solutions of the original system. So do check.
Remark: Alternately, we could write $5\sqrt{s}=2\sqrt{t}+10$ and square both sides. The only reason I did not do that is that $s=(4t-7)/2$, so substitution produces fractions. However, multiplying through by $4$ clears the fractions. In principle the details are pretty similar to the details for the approach I chose, with somewhat more unpleasant numbers.