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(C)=(5C+3)(7C+4)
We move all terms to the left:
(C)-((5C+3)(7C+4))=0
We multiply parentheses ..
-((+35C^2+20C+21C+12))+C=0
We calculate terms in parentheses: -((+35C^2+20C+21C+12)), so:We add all the numbers together, and all the variables
(+35C^2+20C+21C+12)
We get rid of parentheses
35C^2+20C+21C+12
We add all the numbers together, and all the variables
35C^2+41C+12
Back to the equation:
-(35C^2+41C+12)
C-(35C^2+41C+12)=0
We get rid of parentheses
-35C^2+C-41C-12=0
We add all the numbers together, and all the variables
-35C^2-40C-12=0
a = -35; b = -40; c = -12;
Δ = b2-4ac
Δ = -402-4·(-35)·(-12)
Δ = -80
Delta is less than zero, so there is no solution for the equation
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