(g-4)(g+1)=(g+2)(g-6)

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Solution for (g-4)(g+1)=(g+2)(g-6) equation:



(g-4)(g+1)=(g+2)(g-6)
We move all terms to the left:
(g-4)(g+1)-((g+2)(g-6))=0
We multiply parentheses ..
(+g^2+g-4g-4)-((g+2)(g-6))=0
We calculate terms in parentheses: -((g+2)(g-6)), so:
(g+2)(g-6)
We multiply parentheses ..
(+g^2-6g+2g-12)
We get rid of parentheses
g^2-6g+2g-12
We add all the numbers together, and all the variables
g^2-4g-12
Back to the equation:
-(g^2-4g-12)
We get rid of parentheses
g^2-g^2+g-4g+4g-4+12=0
We add all the numbers together, and all the variables
g+8=0
We move all terms containing g to the left, all other terms to the right
g=-8

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