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18x^2-9=0
a = 18; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·18·(-9)
Δ = 648
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{2}}{2*18}=\frac{0-18\sqrt{2}}{36} =-\frac{18\sqrt{2}}{36} =-\frac{\sqrt{2}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{2}}{2*18}=\frac{0+18\sqrt{2}}{36} =\frac{18\sqrt{2}}{36} =\frac{\sqrt{2}}{2} $
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