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=-0.4Y^2-3Y+1
We move all terms to the left:
-(-0.4Y^2-3Y+1)=0
We get rid of parentheses
0.4Y^2+3Y-1=0
a = 0.4; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·0.4·(-1)
Δ = 10.6
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{-(3)-\sqrt{10.6}}{2*0.4}=\frac{-3-\sqrt{10.6}}{0.8} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{10.6}}{2*0.4}=\frac{-3+\sqrt{10.6}}{0.8} $
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