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3y^2=30
We move all terms to the left:
3y^2-(30)=0
a = 3; b = 0; c = -30;
Δ = b2-4ac
Δ = 02-4·3·(-30)
Δ = 360
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{10}}{2*3}=\frac{0-6\sqrt{10}}{6} =-\frac{6\sqrt{10}}{6} =-\sqrt{10} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{10}}{2*3}=\frac{0+6\sqrt{10}}{6} =\frac{6\sqrt{10}}{6} =\sqrt{10} $
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