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3c^2-20c+12=0
a = 3; b = -20; c = +12;
Δ = b2-4ac
Δ = -202-4·3·12
Δ = 256
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{256}=16$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16}{2*3}=\frac{4}{6} =2/3 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16}{2*3}=\frac{36}{6} =6 $
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