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=10-3Y-Y^2
We move all terms to the left:
-(10-3Y-Y^2)=0
We get rid of parentheses
Y^2+3Y-10=0
a = 1; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·1·(-10)
Δ = 49
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{49}=7$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*1}=\frac{-10}{2} =-5 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*1}=\frac{4}{2} =2 $
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