2j^2-7j+4=0

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Solution for 2j^2-7j+4=0 equation: 



2j^2-7j+4=0

a = 2; b = -7; c = +4;
Δ = b2-4ac
Δ = -72-4·2·4
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{17}}{2*2}=\frac{7-\sqrt{17}}{4}
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{17}}{2*2}=\frac{7+\sqrt{17}}{4}
$

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