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3y^2-5=0
a = 3; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·3·(-5)
Δ = 60
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*3}=\frac{0-2\sqrt{15}}{6} =-\frac{2\sqrt{15}}{6} =-\frac{\sqrt{15}}{3} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*3}=\frac{0+2\sqrt{15}}{6} =\frac{2\sqrt{15}}{6} =\frac{\sqrt{15}}{3} $
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