(2x^2)-(5x)+1=0

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Solution for (2x^2)-(5x)+1=0 equation: 



(2x^2)-(5x)+1=0

a = 2; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·2·1
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{17}}{2*2}=\frac{5-\sqrt{17}}{4}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{17}}{2*2}=\frac{5+\sqrt{17}}{4}
$

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