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10c^2-21c=4c+6
We move all terms to the left:
10c^2-21c-(4c+6)=0
We get rid of parentheses
10c^2-21c-4c-6=0
We add all the numbers together, and all the variables
10c^2-25c-6=0
a = 10; b = -25; c = -6;
Δ = b2-4ac
Δ = -252-4·10·(-6)
Δ = 865
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{865}}{2*10}=\frac{25-\sqrt{865}}{20} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{865}}{2*10}=\frac{25+\sqrt{865}}{20} $
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