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0.55s^2+s=0
a = 0.55; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·0.55·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*0.55}=\frac{-2}{1.1} =-1+0.9/1.1 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*0.55}=\frac{0}{1.1} =0 $
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