j(4j-5)=0

Solution for j(4j-5)=0 equation:

j(4j-5)=0

We multiply parentheses

4j^2-5j=0

a = 4; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·4·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*4}=\frac{0}{8}
=0
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*4}=\frac{10}{8}
=1+1/4
$