Hi Louis,
Making your substitutions,
y==t-DT /.DT->300a/9.81
y==-30.58103975535168` a+t/.a->((B/2)^2/3)/((Tu/2)^2/3*(h-hv+tu))
y==t-(30.58103975535168` B^2)/((h-hv+tu) Tu^2)/.hv->(Vi*d)/(4B*(Tu/2)^2)^1/3
y==t-(30.58103975535168` B^2)/(Tu^2 (h+tu-(d Vi)/(3 B Tu^2)))/.B->Vi*g*d
y==t-(30.58103975535168` d^2 g^2 Vi^2)/((h+tu-1/(3 g Tu^2)) Tu^2)/.g->(9.81 (t-x))/300
y==t-(0.032700000000000014` d^2 Vi^2 (t-x)^2)/(Tu^2 (h+tu-10.193679918450558`/(Tu^2 (t-x))))/.Vi->Q/l
y==t-(0.032700000000000014` d^2 Q^2 (t-x)^2)/(l^2 Tu^2 (h+tu-10.193679918450558`/(Tu^2 (t-x))))
So, the final form of y is:
y==t-(A (t-x)^2)/(B-F/(t-x))
This cannot be put into a quadratic form. If you solve for x, there would be two solutions if it were quadratic:
In[21]:= Solve[y==t-(A (t-x)^2)/(B-F/(t-x)),x]
Out[21]= {
{x->t+(2^(1/3) (-3 A B t+3 A B y))/(3 A (-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3))-(1/(3 2^(1/3) A))((-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3))},
{x->t-((1+I Sqrt[3]) (-3 A B t+3 A B y))/(3 2^(2/3) A (-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3))+(1/(6 2^(1/3) A))(1-I Sqrt[3]) (-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3)},
{x->t-((1-I Sqrt[3]) (-3 A B t+3 A B y))/(3 2^(2/3) A (-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3))+(1/(6 2^(1/3) A))(1+I Sqrt[3]) (-27 A^2 F t+27 A^2 F y+\[Sqrt](4 (-3 A B t+3 A B y)^3+(-27 A^2 F t+27 A^2 F y)^2))^(1/3)}
}
But there are three solutions.
Eric