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2=3y^2=19
We move all terms to the left:
2-(3y^2)=0
a = -3; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-3)·2
Δ = 24
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-3}=\frac{0-2\sqrt{6}}{-6} =-\frac{2\sqrt{6}}{-6} =-\frac{\sqrt{6}}{-3} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-3}=\frac{0+2\sqrt{6}}{-6} =\frac{2\sqrt{6}}{-6} =\frac{\sqrt{6}}{-3} $
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