j^2+5j+4=0

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Solution for j^2+5j+4=0 equation: 



j^2+5j+4=0

a = 1; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·1·4
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*1}=\frac{-8}{2}
=-4
$


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

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