# Solve by factorization 6x^2+x-2=0

Solve by factorization
$6{x}^{2}+x-2=0$
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Step 1
factor the following $6{x}^{2}+x-2=0$
Step 2
Factor the quadratic $6{x}^{2}+x-2$
The coefficient of ${x}^{2}$ is 6 and the constant term is −2.
The product of 6 and −2 is −12. The factors of −12 which sum to
1 are −3 and 4
So $6{x}^{2}+x-2=6{x}^{2}+4x-3x-2=2\left(2x-1\right)+3x\left(2x-1\right)$:
2(2x−1)+3x(2x−1):
Factor 2x−1 from 2(2x−1)+3x(2x−1)=(2x−1)(3x+2),
=(2x−1)(3x+2)
Now set (2x−1)(3x+2)=0
(2x−1)=0 and (3x+2)=0
$x=\frac{1}{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x=-\frac{2}{3}$