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Mathematics 12 Online
OpenStudy (anonymous):

solve for c. c^2+7c+12=0

OpenStudy (anonymous):

c^2+4c+3c+12=0 c(c+4)+3(c+4)=0 (c+3)(c+4)=0 c=-3 and -4

OpenStudy (anonymous):

it was coreect

OpenStudy (amistre64):

same steps; trying to find a value for, in this case i spose: c^2 +(a+b)c + (ab) ab = 12 a+b = 7 4 and 3 good; (c+4)(c+3); when c=-4 this = 0; when c=-3, this also =0

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