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-(192x)^2+18x+18=0
a = -192; b = 18; c = +18;
Δ = b2-4ac
Δ = 182-4·(-192)·18
Δ = 14148
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{14148}=\sqrt{36*393}=\sqrt{36}*\sqrt{393}=6\sqrt{393}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{393}}{2*-192}=\frac{-18-6\sqrt{393}}{-384} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{393}}{2*-192}=\frac{-18+6\sqrt{393}}{-384} $
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