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6c(2c+1)=0
We multiply parentheses
12c^2+6c=0
a = 12; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·12·0
Δ = 36
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{36}=6$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*12}=\frac{-12}{24} =-1/2 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*12}=\frac{0}{24} =0 $
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