0=4c^2+6c

Solution for 0=4c^2+6c equation:

0=4c^2+6c

We move all terms to the left:

0-(4c^2+6c)=0

We add all the numbers together, and all the variables

-(4c^2+6c)=0

We get rid of parentheses

-4c^2-6c=0

a = -4; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-4)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*-4}=\frac{0}{-8}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-4}=\frac{12}{-8}
=-1+1/2
$