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2c^2-13c+6=0
a = 2; b = -13; c = +6;
Δ = b2-4ac
Δ = -132-4·2·6
Δ = 121
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{121}=11$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*2}=\frac{2}{4} =1/2 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*2}=\frac{24}{4} =6 $
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