j2+6j+5=0

Solution for j2+6j+5=0 equation:

j2+6j+5=0

We add all the numbers together, and all the variables

j^2+6j+5=0

a = 1; b = 6; c = +5;
Δ = b2-4ac
Δ = 62-4·1·5
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4}{2*1}=\frac{-10}{2}
=-5
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*1}=\frac{-2}{2}
=-1
$