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On line Solution of Monovariate Equation:
    Input any unary equation directly, and then click the "Next" button to obtain the solution of the equation.
    It supports equations that contain mathematical functions.
    Current location:Equations > Monovariate Equation > The history of univariate equation calculation > Answer
    Overview: 1 questions will be solved this time.Among them
           ☆1 equations

[ 1/1 Equation]
    Work: Find the solution of equation (1+2y)(1+2y)-3(1+2y)+2 = 0 .
    Question type: Equation
    Solution:Original question:
     (1 + 2 y )(1 + 2 y )3(1 + 2 y ) + 2 = 0
    Remove the bracket on the left of the equation:
     Left side of the equation = 1(1 + 2 y ) + 2 y (1 + 2 y )3(1 + 2 y ) + 2
                                             = 1 × 1 + 1 × 2 y + 2 y (1 + 2 y )3(1 + 2 y ) + 2
                                             = 1 + 2 y + 2 y (1 + 2 y )3(1 + 2 y ) + 2
                                             = 3 + 2 y + 2 y (1 + 2 y )3(1 + 2 y )
                                             = 3 + 2 y + 2 y × 1 + 2 y × 2 y 3(1 + 2 y )
                                             = 3 + 2 y + 2 y + 4 y y 3(1 + 2 y )
                                             = 3 + 4 y + 4 y y 3(1 + 2 y )
                                             = 3 + 4 y + 4 y y 3 × 13 × 2 y
                                             = 3 + 4 y + 4 y y 36 y
                                             = 02 y + 4 y y
    The equation is transformed into :
     02 y + 4 y y = 0

    After the equation is converted into a general formula, it is converted into:
    ( y - 0 )( 2y - 1 )=0
    From
        y - 0 = 0
        2y - 1 = 0

    it is concluded that::
        y1=0
        y2=
1
2
    
    There are 2 solution(s).


解一元二次方程的详细方法请参阅:《一元二次方程的解法》



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