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3c^2-5c=0
a = 3; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·3·0
Δ = 25
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{25}=5$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*3}=\frac{0}{6} =0 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*3}=\frac{10}{6} =1+2/3 $
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