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=-Y(Y-3)
We move all terms to the left:
-(-Y(Y-3))=0
We calculate terms in parentheses: -(-Y(Y-3)), so:We get rid of parentheses
-Y(Y-3)
We multiply parentheses
-Y^2+3Y
We add all the numbers together, and all the variables
-1Y^2+3Y
Back to the equation:
-(-1Y^2+3Y)
1Y^2-3Y=0
We add all the numbers together, and all the variables
Y^2-3Y=0
a = 1; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·1·0
Δ = 9
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*1}=\frac{0}{2} =0 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*1}=\frac{6}{2} =3 $
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