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0.01X^2+X-5000=0
a = 0.01; b = 1; c = -5000;
Δ = b2-4ac
Δ = 12-4·0.01·(-5000)
Δ = 201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{201}}{2*0.01}=\frac{-1-\sqrt{201}}{0.02} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{201}}{2*0.01}=\frac{-1+\sqrt{201}}{0.02} $
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