## Algebra and Trigonometry 10th Edition

$\left\{\dfrac{686-196\sqrt 6}{25},\dfrac{686+196\sqrt 6}{25}\right\}$
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: $\left\{\dfrac{686-196\sqrt 6}{25},\dfrac{686+196\sqrt 6}{25}\right\}$