Let us generalize a problem, the original problem is Solve for x:
\(\sqrt{3x^2 - 4x + 34} + \sqrt{3x^2 - 4x - 11} = 9\)

Now we generlize this problem: if \(a,b,c_1,c_2,c_3 \isin R\) with a condition that \((ax^2 + bx + c_1),(ax^2 + bx + c_2) \ge 0\) and \(c_3 \gt 0\) and \(\sqrt{ax^2 + bx + c_1} + \sqrt{ax^2 + bx + c_2} = c_3\), then find the value of x

Solution:

Clearly
\((ax^2 + bx + c_1) - (ax^2 + bx + c_2) = c_1 - c_2\)
\(\implies \sqrt{ax^2 + bx + c_1} - \sqrt{ax^2 + bx + c_2} = \tfrac{c_1 - c_2}{c_3}\)

Adding this equation with our original equation, we get

\(\implies 2\sqrt{ax^2 + bx + c_1} = c_3 + \tfrac{c_1-c_2}{c_3}\)
\(\implies 2\sqrt{ax^2 + bx + c_1} = \tfrac{c_3^2 + c_1 - c_2}{c_3}\)
\(\implies \sqrt{ax^2 + bx + c_1} = \tfrac{c_3^2 + c_1 - c_2}{2c_3}\)
\(\implies ax^2 + bx + c_1 = (\tfrac{c_3^2 + c_1 - c_2}{2c_3})^2\)
And finally we can solve for x, which is left as an exercise for the reader! The expressions above are a bit complex and so we can just leave this and instead we can say
\(\sqrt{ax^2 + bx + c_1} = \tfrac{c_3^2 + c_1 - c_2}{2c_3}\)
Now coming to the original problem,
we have \(a=3, b=-4, c_1 = 34, c_2 = -11, c_3 = 9\) and we'll have

\(3x^2 - 4x - 15 = 0\)
\(\implies x = 3,-5/3\)

In similar way we can generalize for:
\(\sqrt{ax^2 + bx + c_1} - \sqrt{ax^2 + bx + c_2} = c_3\)