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=2Y^2-12Y-40
We move all terms to the left:
-(2Y^2-12Y-40)=0
We get rid of parentheses
-2Y^2+12Y+40=0
a = -2; b = 12; c = +40;
Δ = b2-4ac
Δ = 122-4·(-2)·40
Δ = 464
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{29}}{2*-2}=\frac{-12-4\sqrt{29}}{-4} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{29}}{2*-2}=\frac{-12+4\sqrt{29}}{-4} $
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