2x^2-19x+19=0

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Solution for 2x^2-19x+19=0 equation:



2x^2-19x+19=0
a = 2; b = -19; c = +19;
Δ = b2-4ac
Δ = -192-4·2·19
Δ = 209
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{-(-19)-\sqrt{209}}{2*2}=\frac{19-\sqrt{209}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{209}}{2*2}=\frac{19+\sqrt{209}}{4} $

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