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c^2-15c+14=0
a = 1; b = -15; c = +14;
Δ = b2-4ac
Δ = -152-4·1·14
Δ = 169
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{169}=13$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-13}{2*1}=\frac{2}{2} =1 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+13}{2*1}=\frac{28}{2} =14 $
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