Question

Помогиитееее 2у квадрате2 +3у+1=0

Answer 1

РЕШЕНИЕ:

${2y^2+3y+1 = 0} \\\\\\D = b^2-4ac \\ \\D = 3^2-4\cdot2\cdot 1 =9-8 = 1 = 1^2\\\\D > 0\\\\\\x_1_,_2=\dfrac{-b\pm\sqrt{D}}{2a}\\\\\\x_1 = \dfrac{-3+1}{2\cdot2} = \dfrac{2}{4} = \dfrac{1}{2} = -0.5\\\\\\x_2 = \dfrac{-3-1}{2\cdot2} = \dfrac{-4}{4} = -1$

ОТВЕТ: -1; -0,5