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18=3b^2
We move all terms to the left:
18-(3b^2)=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:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*-3}=\frac{0-6\sqrt{6}}{-6} =-\frac{6\sqrt{6}}{-6} =-\frac{\sqrt{6}}{-1} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*-3}=\frac{0+6\sqrt{6}}{-6} =\frac{6\sqrt{6}}{-6} =\frac{\sqrt{6}}{-1} $
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