Question

Реши уровнения
СРОЧНО ПЖПЖПЖПЖПЖ
1)y в квадрате -3y-4=0
2)2y в квадрате -3y+1=0
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Answer 1

$\displaystyle\bf\\1)\\\\y^{2} -3y-4=0\\\\D=(-3)^{2} -4\cdot (-4)=9+16=25=5^{2} \\\\\\y_{1} =\frac{3-5}{2} =\frac{-2}{2} =-1\\\\\\y_{2} =\frac{3+5}{2} =\frac{8}{2} =4\\\\\\Otvet \ . \ -1 \ ; \ 4\\\\\\2)\\\\2y^{2} -3y+1=0\\\\D=(-3)^{2} -4\cdot 2\cdot 1=9-8=1\\\\\\y_{1} =\frac{3-1}{2\cdot 2} =\frac{2}{4} =\frac{1}{2} =0,5\\\\\\y_{2} =\frac{3+1}{2\cdot 2}=\frac{4}{4} =1\\\\\\Otvet \ : \ 0,5 \ ; \ 1$