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5c^2-16c+3=0
a = 5; b = -16; c = +3;
Δ = b2-4ac
Δ = -162-4·5·3
Δ = 196
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{196}=14$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-14}{2*5}=\frac{2}{10} =1/5 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+14}{2*5}=\frac{30}{10} =3 $
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