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b^2+5b-168=0
a = 1; b = 5; c = -168;
Δ = b2-4ac
Δ = 52-4·1·(-168)
Δ = 697
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{-(5)-\sqrt{697}}{2*1}=\frac{-5-\sqrt{697}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{697}}{2*1}=\frac{-5+\sqrt{697}}{2} $
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