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3a^2=18
We move all terms to the left:
3a^2-(18)=0
a = 3; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·3·(-18)
Δ = 216
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*3}=\frac{0-6\sqrt{6}}{6} =-\frac{6\sqrt{6}}{6} =-\sqrt{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*3}=\frac{0+6\sqrt{6}}{6} =\frac{6\sqrt{6}}{6} =\sqrt{6} $
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