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(3c+5)(6c-8)=0
We multiply parentheses ..
(+18c^2-24c+30c-40)=0
We get rid of parentheses
18c^2-24c+30c-40=0
We add all the numbers together, and all the variables
18c^2+6c-40=0
a = 18; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·18·(-40)
Δ = 2916
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{2916}=54$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-54}{2*18}=\frac{-60}{36} =-1+2/3 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+54}{2*18}=\frac{48}{36} =1+1/3 $
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