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Question

Solve each of the following quadratic equations: 
3x+112=23x1,x1,13


Solution

3x+112=23x1

3x+123x1=12

[3(3x1)2(x+1)](x+1)(3x1)=12

(9x32x2)(3x2x+3x1)=12

2(7x5)=(3x2+2x1)

3x2+2x114x+10=0

3x212x+9=0

x24x+3=0

x23xx+3=0

x(x3)1(x3)=0

(x3)(x1)=0

Either 

x3=0

x=3

or

x1=0

x=1

So x = 1, 3
 

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