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3y^2-y2=27
We move all terms to the left:
3y^2-y2-(27)=0
We add all the numbers together, and all the variables
2y^2-27=0
a = 2; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·2·(-27)
Δ = 216
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*2}=\frac{0-6\sqrt{6}}{4} =-\frac{6\sqrt{6}}{4} =-\frac{3\sqrt{6}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*2}=\frac{0+6\sqrt{6}}{4} =\frac{6\sqrt{6}}{4} =\frac{3\sqrt{6}}{2} $
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