Question

Urgente!!! Preciso muito de ajuda nessa equação de SEGUNDO grau pooor favooor!!

$\frac{y-1}{2}- \frac{y(3-y)}{3}= \frac{y+1}{3}$

Answer 1

$\dfrac{y-1}{2}-\dfrac{y(3-y)}{3}=\dfrac{y+1}{3}\\\\ 3(y-1)-2y(3-y)=2(y+1)\\\\ 3y-3-6y+2y^2=2y+2\\\\ 2y^2-5y-5=0\\\\ \Delta=b^2-4\cdot a\cdo c\\ \Delta=(-5)^2-4\cdot2\cdot(-5)\\ \Delta=25+40\\ \Delta=65\\\\ y=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{5\pm\sqrt{65}}{2\cdot2}=\dfrac{5\pm\sqrt{65}}{4}\\\\ y_1=\dfrac{5+\sqrt{65}}{4}\;\;\;\;\;\;\;\;\;\;y_2=\dfrac{5-\sqrt{65}}{4}\\\\ S=\{\dfrac{5-\sqrt{65}}{4},\dfrac{5+\sqrt{65}}{4}\}$