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b^2-6b-160=0
a = 1; b = -6; c = -160;
Δ = b2-4ac
Δ = -62-4·1·(-160)
Δ = 676
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{676}=26$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-26}{2*1}=\frac{-20}{2} =-10 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+26}{2*1}=\frac{32}{2} =16 $
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