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

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There are probably a couple of ways to do this, but I chose multiplying it out in polynomial form to start with.

(2z-2)(z+3)=0 ( I think it's the distributive law(I know it's one of the laws...) for the multiplication that's done, to make it a polynomial.)

2z(z) +(2z(3)) -2z-6=0

(^2, this ^denotes raised to the power of, in this case 2, so z^2 is z squared)

2z^2+6z-2z+6=0 simplify

2z^2+4z-6=0

z^2+2z-3=0

(z+3)(z-1)=0 now from here it's in a more familiar and in an easier form to resolve

z+3=0 and z-1=0

z+3=0

z= -3

now we solve for

z-1=0

z=1

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