# r^2-50r+200=0

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## Solution for r^2-50r+200=0 equation:

r^2-50r+200=0
a = 1; b = -50; c = +200;Δ = b2-4acΔ = -502-4·1·200Δ = 1700The delta value is higher than zero, so the equation has two solutionsWe use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$The end solution:
$\sqrt{\Delta}=\sqrt{1700}=\sqrt{100*17}=\sqrt{100}*\sqrt{17}=10\sqrt{17}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-10\sqrt{17}}{2*1}=\frac{50-10\sqrt{17}}{2}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+10\sqrt{17}}{2*1}=\frac{50+10\sqrt{17}}{2}$

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