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x^2-250x+2500=0
a = 1; b = -250; c = +2500;
Δ = b2-4ac
Δ = -2502-4·1·2500
Δ = 52500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52500}=\sqrt{2500*21}=\sqrt{2500}*\sqrt{21}=50\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-250)-50\sqrt{21}}{2*1}=\frac{250-50\sqrt{21}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-250)+50\sqrt{21}}{2*1}=\frac{250+50\sqrt{21}}{2} $
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