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c(11+6c)=35
We move all terms to the left:
c(11+6c)-(35)=0
We add all the numbers together, and all the variables
c(6c+11)-35=0
We multiply parentheses
6c^2+11c-35=0
a = 6; b = 11; c = -35;
Δ = b2-4ac
Δ = 112-4·6·(-35)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{961}=31$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-31}{2*6}=\frac{-42}{12} =-3+1/2 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+31}{2*6}=\frac{20}{12} =1+2/3 $
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