Answer
$\left\{\dfrac{686-196\sqrt 6}{25},\dfrac{686+196\sqrt 6}{25}\right\}$
Work Step by Step
We are given the equation:
$\left(\dfrac{5}{7}x-14\right)^2=8x$
Put the equation in standard form:
$\dfrac{25}{49}x^2-20x+196=8x$
$\dfrac{25}{49}x^2-20x+196-8x=0$
$\dfrac{25}{49}x^2-28x+196=0$
Multiply all terms by 49 to clear denominators:
$49\cdot\dfrac{25}{49}x^2-49\cdot 28x+49\cdot 196=49\cdot 0$
$25x^2-1372x+9604=0$
Solve the equation using the quadratic formula:
$x=\dfrac{-(-1372)\pm\sqrt{(-1372)^2-4(25)(9604)}}{2(25)}=\dfrac{1372\pm\sqrt{921,984}}{50}$
$=\dfrac{1372\pm 392\sqrt 6}{50}$
$=\dfrac{686\pm 196\sqrt 6}{25}$
$x_1=\dfrac{686-196\sqrt 6}{25}$
$x_2=\dfrac{686+196\sqrt 6}{25}$
The solution set of the equation is:
$\left\{\dfrac{686-196\sqrt 6}{25},\dfrac{686+196\sqrt 6}{25}\right\}$
