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(3c)(3c-4)=0
We multiply parentheses
9c^2-12c=0
a = 9; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·9·0
Δ = 144
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{144}=12$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*9}=\frac{0}{18} =0 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*9}=\frac{24}{18} =1+1/3 $
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