2j^2+5j+2=0

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Solution for 2j^2+5j+2=0 equation: 



2j^2+5j+2=0

a = 2; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·2·2
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*2}=\frac{-8}{4}
=-2
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*2}=\frac{-2}{4}
=-1/2
$

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