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=-0.0003Y^2+0.0559Y-0.4021
We move all terms to the left:
-(-0.0003Y^2+0.0559Y-0.4021)=0
We get rid of parentheses
0.0003Y^2-0.0559Y+0.4021=0
a = 0.0003; b = -0.0559; c = +0.4021;
Δ = b2-4ac
Δ = -0.05592-4·0.0003·0.4021
Δ = 0.00264229
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.0559)-\sqrt{0.00264229}}{2*0.0003}=\frac{0.0559-\sqrt{0.00264229}}{0.0006} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.0559)+\sqrt{0.00264229}}{2*0.0003}=\frac{0.0559+\sqrt{0.00264229}}{0.0006} $
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