5c(9c+4)=0

Solution for 5c(9c+4)=0 equation:

5c(9c+4)=0

We multiply parentheses

45c^2+20c=0

a = 45; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·45·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*45}=\frac{-40}{90}
=-4/9
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*45}=\frac{0}{90}
=0
$