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=3-0.5Y^2
We move all terms to the left:
-(3-0.5Y^2)=0
We get rid of parentheses
0.5Y^2-3=0
a = 0.5; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·0.5·(-3)
Δ = 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{-(0)-\sqrt{6}}{2*0.5}=\frac{0-\sqrt{6}}{1} =-\frac{\sqrt{}}{1} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{6}}{2*0.5}=\frac{0+\sqrt{6}}{1} =\frac{\sqrt{}}{1} $
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