Finesse vs. brute force in solving quadratic equations In Higher Algebra by Hall and Knight, the following "artifice" for solving a certain type of  equations is given:

Solve: $\sqrt{3x^2-4x+34} - \sqrt{3x^2-4x-11} = 9$

They make use of the fact that $(3x^2-4x+34) - (3x^2-4x-11) = 45$, and utilizing the formula $a^2 - b^2 = (a-b)(a+b)$, obtain a neat answer that $x = 3, -5/3$.
My attempt, however, results in something weird:
$\sqrt{3x^2-4x+34} - \sqrt{3x^2-4x+11} = 9$
Let $3x^2 - 4x = t$
$\Rightarrow \sqrt{t+34} + \sqrt{t+11}$ = 9
Squaring both sides,
$t + 34 + t + 11 + 2 \sqrt{(t+34)(t+11)} = 81$
$\Rightarrow 18-t = \sqrt{(t+34)(t+11)}$
Squaring again,
$t^2 - 36t + 324 = t^2 + 45t + 374$
$\Rightarrow t = -50/81$
$\Rightarrow 3x^2 - 4x = -50/81$
$\Rightarrow 243x^2 - 324x + 50 = 0$
The discriminant here is $\sqrt{324^2 - 200.243} = 56376$, which is not a perfect square and hence the roots are irrational.
Have I made a mistake somewhere or it is really the case that finesse leads to correct answers?

Assuming the real, actual question has a plus sign between the square roots and $\,-11\,$ in the second square root, and doing your substitution:
$$\sqrt{3x^2-4x+34}+\sqrt{3x^2-4x-11}=9\;,\;\;\color{red}{t:=3x^2-4x}\implies$$
$$\sqrt{t+34}+\sqrt{t-11}=9\implies 2t+23+2\sqrt{(t+34)(t-11)}=81\implies$$
$$29-t=\sqrt{(t+34)(t-11)}\implies t^2-58t+841=t^2+23t-374\implies$$
$$81t=1,215\implies t=15\implies$$
$$15=\color{red}{t=3x^2-4x}\implies 0=3x^2-4x-15=3(x-3)\left(x+\frac53\right)$$
