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=2Y^2+12Y+14
We move all terms to the left:
-(2Y^2+12Y+14)=0
We get rid of parentheses
-2Y^2-12Y-14=0
a = -2; b = -12; c = -14;
Δ = b2-4ac
Δ = -122-4·(-2)·(-14)
Δ = 32
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{2}}{2*-2}=\frac{12-4\sqrt{2}}{-4} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{2}}{2*-2}=\frac{12+4\sqrt{2}}{-4} $
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