Question

2x²+x-6=0 find the roots of the following question​

Answer 1

2x^2 + x - 6 = 0

=> 2x^2 +4x - 3x - 6 = 0

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

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

=> x = -2 (or) x = 3/2

∴ The roots of the given quadratic equation = -2 , 3/2

Answer 2

Your question :

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

Solution:

$2 {x}^{2} + x - 6 = 0\\ 2 {x}^{2} + (4 - 3)x - 6 = 0 \\ 2 {x}^{2} + 4x - 3x - 6 = 0 \\ 2x(x + 2) - 3(x + 2) = 0 \\ (2x - 3)(x + 2) = 0 \\ = > 2x - 3 = 0 \: \: \: x + 2 = 0$

Taking 2x-3 we get,

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

Again taking x+2 =0 we get,

$x + 2 = 0 \\ x = - 2$

Therefore the roots of the equation are

x=-2 or x=3/2