How do you solve $2x(x - 1) = 3$ ?

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Answer 1 · 2016-03-26T06:08:14

$x=[1+-sqrt(13)]/4$

$2x(x-1)=3$

$4x^2-2x=3$

$4x^2-2x-3=0$

Comparing with $ax^2+bx+c=0$ we get,

$a=4, b=-2,c=-3$

By Formula method,

$x=[-b+-sqrt(b^2-4ac)]/(2a)$

$x=[-(-2)+-sqrt((-2)^2-4xx4xx-(3))]/(2xx4)$

$x=[2+-sqrt(4+48)]/(8)$

$x=[2+-sqrt(52)]/(8)$

$x=[2+-sqrt(4xx13)]/(8)$

$x=[2+-2sqrt(13)]/(8)$

$x=(2[1+-sqrt(13)])/8$

$x=[1+-sqrt(13)]/4$

How do you solve 2x(x - 1) = 3 2x(x - 1) = 3?