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=-2Y^2+16Y-1
We move all terms to the left:
-(-2Y^2+16Y-1)=0
We get rid of parentheses
2Y^2-16Y+1=0
a = 2; b = -16; c = +1;
Δ = b2-4ac
Δ = -162-4·2·1
Δ = 248
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{62}}{2*2}=\frac{16-2\sqrt{62}}{4} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{62}}{2*2}=\frac{16+2\sqrt{62}}{4} $
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