2c(c-5)=0

Solution for 2c(c-5)=0 equation:

2c(c-5)=0

We multiply parentheses

2c^2-10c=0

a = 2; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·2·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*2}=\frac{0}{4}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*2}=\frac{20}{4}
=5
$