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25x^2-350x+229=0
a = 25; b = -350; c = +229;
Δ = b2-4ac
Δ = -3502-4·25·229
Δ = 99600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{99600}=\sqrt{400*249}=\sqrt{400}*\sqrt{249}=20\sqrt{249}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-350)-20\sqrt{249}}{2*25}=\frac{350-20\sqrt{249}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-350)+20\sqrt{249}}{2*25}=\frac{350+20\sqrt{249}}{50} $
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