Solve the following quadratic equations by factorization:

We have been given

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

Now we solve the above quadratic equation.

Therefore,

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

$2 x(3 x-2)+1(3 x-2)=0$

$(2 x+1)(3 x-2)=0$

Now, one of the products must be equal to zero for the whole product to be zero. Hence we equate both the products to zero in order to find the value of x.

Therefore,

$2 x+1=0$

$2 x=-1$

$x=\frac{-1}{2}$

Or

$3 x-2=0$

$3 x=2$

$x=\frac{2}{3}$

Hence, $x=\frac{-1}{2}$ or $x=\frac{2}{3}$.