6c^2+12c^2-18c=0

Solution for 6c^2+12c^2-18c=0 equation:

6c^2+12c^2-18c=0

We add all the numbers together, and all the variables

18c^2-18c=0

a = 18; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·18·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*18}=\frac{0}{36}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*18}=\frac{36}{36}
=1
$