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x^2-250x+5=0
a = 1; b = -250; c = +5;
Δ = b2-4ac
Δ = -2502-4·1·5
Δ = 62480
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{62480}=\sqrt{16*3905}=\sqrt{16}*\sqrt{3905}=4\sqrt{3905}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-250)-4\sqrt{3905}}{2*1}=\frac{250-4\sqrt{3905}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-250)+4\sqrt{3905}}{2*1}=\frac{250+4\sqrt{3905}}{2} $
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