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8-104+2y^2=16
We move all terms to the left:
8-104+2y^2-(16)=0
We add all the numbers together, and all the variables
2y^2-112=0
a = 2; b = 0; c = -112;
Δ = b2-4ac
Δ = 02-4·2·(-112)
Δ = 896
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{14}}{2*2}=\frac{0-8\sqrt{14}}{4} =-\frac{8\sqrt{14}}{4} =-2\sqrt{14} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{14}}{2*2}=\frac{0+8\sqrt{14}}{4} =\frac{8\sqrt{14}}{4} =2\sqrt{14} $
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