Question

How many solutions will the following equation have? 25=20x-4x^2

Answer 1

Hello,

$4x^2-20x+25=0\\ \Delta=20^2-4*4*25=0\\ x= \dfrac{20}{8} = \dfrac{5}{2} \\ 4x^2-20x+25=0\\ (2x-5)^2=0\\ \mbox{One solution (double) Sol=\{ \dfrac{5}{2} \}}$