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18x^2-193x+492=0
a = 18; b = -193; c = +492;
Δ = b2-4ac
Δ = -1932-4·18·492
Δ = 1825
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-193)-5\sqrt{73}}{2*18}=\frac{193-5\sqrt{73}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-193)+5\sqrt{73}}{2*18}=\frac{193+5\sqrt{73}}{36} $
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