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x^2+0.0020x-(4.9)10^-5=0
We add all the numbers together, and all the variables
x^2+0.0020x=0
a = 1; b = 0.0020; c = 0;
Δ = b2-4ac
Δ = 0.00202-4·1·0
Δ = 4.0E-6
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{-(0.0020)-\sqrt{4.0E-6}}{2*1}=\frac{-0.002-\sqrt{4.0E-6}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.0020)+\sqrt{4.0E-6}}{2*1}=\frac{-0.002+\sqrt{4.0E-6}}{2} $
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