Answer
$x=\dfrac{17\pm \sqrt{145}}{9}$
Work Step by Step
Using the properties of equality, the given quadratic equation, $
(3x-4)^2=10x
,$ is equivalent to
\begin{array}{l}\require{cancel}
(3x)^2+2(3x)(-4)+(-4)^2=10x
\\\\
9x^2-24x+16=10x
\\\\
9x^2+(-24x-10x)+16=0
\\\\
9x^2-34x+16=0
.\end{array}
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solutions of the quadratic equation above are
\begin{array}{l}\require{cancel}
x=\dfrac{-(-34)\pm\sqrt{(-34)^2-4(9)(16)}}{2(9)}
\\\\
x=\dfrac{34\pm\sqrt{1156-576}}{18}
\\\\
x=\dfrac{34\pm\sqrt{580}}{18}
\\\\
x=\dfrac{34\pm\sqrt{4\cdot145}}{18}
\\\\
x=\dfrac{34\pm\sqrt{(2)^2\cdot145}}{18}
\\\\
x=\dfrac{34\pm2\sqrt{145}}{18}
\\\\
x=\dfrac{2(17\pm \sqrt{145})}{18}
\\\\
x=\dfrac{\cancel{2}(17\pm \sqrt{145})}{\cancel{2}(9)}
\\\\
x=\dfrac{17\pm \sqrt{145}}{9}
.\end{array}
