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-3c=-7c^2
We move all terms to the left:
-3c-(-7c^2)=0
We get rid of parentheses
7c^2-3c=0
a = 7; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·7·0
Δ = 9
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{9}=3$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*7}=\frac{0}{14} =0 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*7}=\frac{6}{14} =3/7 $
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