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c^2-27c+150=0
a = 1; b = -27; c = +150;
Δ = b2-4ac
Δ = -272-4·1·150
Δ = 129
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-\sqrt{129}}{2*1}=\frac{27-\sqrt{129}}{2} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{129}}{2*1}=\frac{27+\sqrt{129}}{2} $
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