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=2Y^2+12Y+16
We move all terms to the left:
-(2Y^2+12Y+16)=0
We get rid of parentheses
-2Y^2-12Y-16=0
a = -2; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·(-2)·(-16)
Δ = 16
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{16}=4$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4}{2*-2}=\frac{8}{-4} =-2 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*-2}=\frac{16}{-4} =-4 $
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