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=2Y^2+5Y-3=Y+27
We move all terms to the left:
-(2Y^2+5Y-3)=0
We get rid of parentheses
-2Y^2-5Y+3=0
a = -2; b = -5; c = +3;
Δ = b2-4ac
Δ = -52-4·(-2)·3
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$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*-2}=\frac{-2}{-4} =1/2 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*-2}=\frac{12}{-4} =-3 $
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