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=-16Y^2+540
We move all terms to the left:
-(-16Y^2+540)=0
We get rid of parentheses
16Y^2-540=0
a = 16; b = 0; c = -540;
Δ = b2-4ac
Δ = 02-4·16·(-540)
Δ = 34560
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{34560}=\sqrt{2304*15}=\sqrt{2304}*\sqrt{15}=48\sqrt{15}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{15}}{2*16}=\frac{0-48\sqrt{15}}{32} =-\frac{48\sqrt{15}}{32} =-\frac{3\sqrt{15}}{2} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{15}}{2*16}=\frac{0+48\sqrt{15}}{32} =\frac{48\sqrt{15}}{32} =\frac{3\sqrt{15}}{2} $
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