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3a^2=324
We move all terms to the left:
3a^2-(324)=0
a = 3; b = 0; c = -324;
Δ = b2-4ac
Δ = 02-4·3·(-324)
Δ = 3888
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3888}=\sqrt{1296*3}=\sqrt{1296}*\sqrt{3}=36\sqrt{3}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36\sqrt{3}}{2*3}=\frac{0-36\sqrt{3}}{6} =-\frac{36\sqrt{3}}{6} =-6\sqrt{3} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36\sqrt{3}}{2*3}=\frac{0+36\sqrt{3}}{6} =\frac{36\sqrt{3}}{6} =6\sqrt{3} $
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