4j(j+8)=0

Solution for 4j(j+8)=0 equation:

4j(j+8)=0

We multiply parentheses

4j^2+32j=0

a = 4; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·4·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*4}=\frac{-64}{8}
=-8
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*4}=\frac{0}{8}
=0
$