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y^2=330
We move all terms to the left:
y^2-(330)=0
a = 1; b = 0; c = -330;
Δ = b2-4ac
Δ = 02-4·1·(-330)
Δ = 1320
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1320}=\sqrt{4*330}=\sqrt{4}*\sqrt{330}=2\sqrt{330}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{330}}{2*1}=\frac{0-2\sqrt{330}}{2} =-\frac{2\sqrt{330}}{2} =-\sqrt{330} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{330}}{2*1}=\frac{0+2\sqrt{330}}{2} =\frac{2\sqrt{330}}{2} =\sqrt{330} $
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