5j^2+5j+1=0

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Solution for 5j^2+5j+1=0 equation: 



5j^2+5j+1=0

a = 5; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·5·1
Δ = 5
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{-(5)-\sqrt{5}}{2*5}=\frac{-5-\sqrt{5}}{10}
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{5}}{2*5}=\frac{-5+\sqrt{5}}{10}
$

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