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=3Y^2-12Y+3
We move all terms to the left:
-(3Y^2-12Y+3)=0
We get rid of parentheses
-3Y^2+12Y-3=0
a = -3; b = 12; c = -3;
Δ = b2-4ac
Δ = 122-4·(-3)·(-3)
Δ = 108
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{3}}{2*-3}=\frac{-12-6\sqrt{3}}{-6} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{3}}{2*-3}=\frac{-12+6\sqrt{3}}{-6} $
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