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500x^2+441x-1764=0
a = 500; b = 441; c = -1764;
Δ = b2-4ac
Δ = 4412-4·500·(-1764)
Δ = 3722481
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3722481}=\sqrt{441*8441}=\sqrt{441}*\sqrt{8441}=21\sqrt{8441}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(441)-21\sqrt{8441}}{2*500}=\frac{-441-21\sqrt{8441}}{1000} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(441)+21\sqrt{8441}}{2*500}=\frac{-441+21\sqrt{8441}}{1000} $
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