6c2-19c+-3=0

Solution for 6c2-19c+-3=0 equation:

6c^2-19c+-3=0

We add all the numbers together, and all the variables

6c^2-19c=0

a = 6; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·6·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*6}=\frac{0}{12}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*6}=\frac{38}{12}
=3+1/6
$