Решите уравнение 50y-10y^2=55

[tex]50y-10y^2=55\\ -10y^{2}+50y-55=0\\ D=b^{2}-4ac=50^{2}-4*(-10)*(-55)=300\\ x_{1}= \frac{-b+\sqrt{D}}{2a} = \frac{-50+\sqrt{300}}{-20}= \frac{5+\sqrt{3}}{2} \\ \\ x_{2}= \frac{-b-\sqrt{D}}{2a} = \frac{-50-\sqrt{300}}{-20}= \frac{5-\sqrt{3}}{2} \\ \\[/tex]