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b^2+27b-140=0
a = 1; b = 27; c = -140;
Δ = b2-4ac
Δ = 272-4·1·(-140)
Δ = 1289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-\sqrt{1289}}{2*1}=\frac{-27-\sqrt{1289}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{1289}}{2*1}=\frac{-27+\sqrt{1289}}{2} $
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