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9u^2-3=0
a = 9; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·9·(-3)
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*9}=\frac{0-6\sqrt{3}}{18} =-\frac{6\sqrt{3}}{18} =-\frac{\sqrt{3}}{3} $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*9}=\frac{0+6\sqrt{3}}{18} =\frac{6\sqrt{3}}{18} =\frac{\sqrt{3}}{3} $
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