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18x^2-63=0
a = 18; b = 0; c = -63;
Δ = b2-4ac
Δ = 02-4·18·(-63)
Δ = 4536
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4536}=\sqrt{324*14}=\sqrt{324}*\sqrt{14}=18\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{14}}{2*18}=\frac{0-18\sqrt{14}}{36} =-\frac{18\sqrt{14}}{36} =-\frac{\sqrt{14}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{14}}{2*18}=\frac{0+18\sqrt{14}}{36} =\frac{18\sqrt{14}}{36} =\frac{\sqrt{14}}{2} $
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