i(i+5)=0.

Solution for i(i+5)=0. equation:

i(i+5)=0.

We multiply parentheses

i^2+5i=0

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

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