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=-16Y^2+32
We move all terms to the left:
-(-16Y^2+32)=0
We get rid of parentheses
16Y^2-32=0
a = 16; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·16·(-32)
Δ = 2048
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{2}}{2*16}=\frac{0-32\sqrt{2}}{32} =-\frac{32\sqrt{2}}{32} =-\sqrt{2} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{2}}{2*16}=\frac{0+32\sqrt{2}}{32} =\frac{32\sqrt{2}}{32} =\sqrt{2} $
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