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9b^2=18
We move all terms to the left:
9b^2-(18)=0
a = 9; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·9·(-18)
Δ = 648
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{2}}{2*9}=\frac{0-18\sqrt{2}}{18} =-\frac{18\sqrt{2}}{18} =-\sqrt{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{2}}{2*9}=\frac{0+18\sqrt{2}}{18} =\frac{18\sqrt{2}}{18} =\sqrt{2} $
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