Math, asked by Anonymous, 6 months ago

Can anyone Solve this...?

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Answered by silentloffer

2

Answer:

$2 {x}^{2} + 3x - 2x - 6 = 0$

$2 {x}^{2} + x - 6 = 0$

$2 {x}^{2} + 4x - 3x - 6 = 0$

$2x(x + 2) - 3(x + 2) = 0$

$(2x - 3)(x + 2) = 0$

$2x - 3 = 0 = = > x = \frac{3}{2} \: ans..$

$x + 2 = 0 = = = > > x = - 2 \: \: ans...$

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Answered by itz123456

0

Answer:

2x²+ 3x-2x-6 =0

2x²+ x -6 =0

2x²+ 4x-3x -6=0

2x(x+2) -3(x+2) =0

(x+2) (2x-3) =0

x+2=0 2x-3=0

x= -2 2x= 3

x= 3/2

Answer: x= -2 , 3/2

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