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10c^2-33c+20=0
a = 10; b = -33; c = +20;
Δ = b2-4ac
Δ = -332-4·10·20
Δ = 289
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{289}=17$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-17}{2*10}=\frac{16}{20} =4/5 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+17}{2*10}=\frac{50}{20} =2+1/2 $
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