(2u^2)-9u+5=0

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Solution for (2u^2)-9u+5=0 equation: 



(2u^2)-9u+5=0

a = 2; b = -9; c = +5;
Δ = b2-4ac
Δ = -92-4·2·5
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{41}}{2*2}=\frac{9-\sqrt{41}}{4}
$


$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{41}}{2*2}=\frac{9+\sqrt{41}}{4}
$

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