Answer
$x_{1}=\dfrac{17 + \sqrt{145}}{9}$ and $x_{2}=\dfrac{17 - \sqrt{145}}{9}$
Work Step by Step
Given $(3x-4)^2=10x \longrightarrow (3x)^2 - 2(3x)(4) + 4^2 = 10x \longrightarrow \\ 9x^2 - 24x + 16 - 10x = 0 \longrightarrow 9x^2 - 34x + 16 = 0$
$a = 9, \ b = -34, \ c = 16$
Using the quadratic formula: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-(-34) \pm \sqrt{(-34)^2-4\times 9\times 16}}{2\times 9} \longrightarrow \dfrac{34\pm \sqrt{1156-576}}{18} = \dfrac{34\pm \sqrt{580}}{18} = \dfrac{34\pm 2\sqrt{145}}{18} = \dfrac{17\pm \sqrt{145}}{9}$
Therefore, the solutions are $x_{1}=\dfrac{17 + \sqrt{145}}{9}$ and $x_{2}=\dfrac{17 - \sqrt{145}}{9}$
