Solve the equation.

(2r + 1 )( r -2 ) = - (3r -6 )

Solve the equation.

(2r + 1 )( r -2 ) = - (3r -6 )

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*I like answers above, but want to give another approach, this is algebra any way.*

(2r + 1)(r - 2) = - (3r - 6)

(2r + 1)(r - 2) + (3r - 6) = 0

(2r + 1)(r - 2) + 3(r - 2) = 0 , factor out (r - 2)

(r - 2)(2r + 1 + 3) = 0

(r - 2)(2r + 4) = 0 , factor out 2

2(r - 2)(r + 2) = 0

Product equal zero if at least one factor equal zero.

r_{1,2} = ±2

(2r + 1)(r - 2) = -(3r - 6)

2r^{2 }- 4r + r - 2 = -3r + 6

2r^{2 }- 3r - 2 = -3r + 6

2r^{2 }- 3r + 3r - 2 - 6 = 0

2r^{2 }- 8 = 0

(2r^{2} - 8) / 2 = 0 / 2

r^{2 }- 4 = 0

r^{2 }= 4

**r = ± 2**

Expand and simplify,

2r^{2} - 3r - 2 = -3r + 6

r^{2} = 4

r = +/- 2 <==Answer

(2r + 1 )( r -2 ) = - (3r -6 )

2r^{2}+ r-4r-2 = -3r+6

2r^{2}-3r-2=-3r+6

2r^{2}=8

r^{2}=4

Thus r= +2 or -2

2r^2 + r - 4r -2 + 3r -6 = 0 = 2r^2 - 8 = (2r + 4)(r - 2)

r = 2 or -2

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