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=-3Y^2+1
We move all terms to the left:
-(-3Y^2+1)=0
We get rid of parentheses
3Y^2-1=0
a = 3; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·3·(-1)
Δ = 12
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*3}=\frac{0-2\sqrt{3}}{6} =-\frac{2\sqrt{3}}{6} =-\frac{\sqrt{3}}{3} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*3}=\frac{0+2\sqrt{3}}{6} =\frac{2\sqrt{3}}{6} =\frac{\sqrt{3}}{3} $
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