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5x^2-25x+10=0
a = 5; b = -25; c = +10;
Δ = b2-4ac
Δ = -252-4·5·10
Δ = 425
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{17}}{2*5}=\frac{25-5\sqrt{17}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{17}}{2*5}=\frac{25+5\sqrt{17}}{10} $
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