Answer
$x=\displaystyle \frac{2}{5}$
Work Step by Step
$ 25x^{2}-20x+40\qquad$... solve with the Quadractic formula. $a=25,\ b=-20,\ c=4$
$ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-20,\ a$ for $25$ and $c$ for $4$.
$ x=\displaystyle \frac{-(-20)\pm\sqrt{(-20)^{2}-4\cdot(4)\cdot 25}}{2\cdot 25}\qquad$... simplify.
$x=\displaystyle \frac{20\pm\sqrt{400-400}}{50}$
$x=\displaystyle \frac{20\pm\sqrt{0}}{50}$
$x=\displaystyle \frac{20}{50}$
$x=\displaystyle \frac{2}{5}$
