0=3c2-45c+42

Solution for 0=3c2-45c+42 equation:

0=3c^2-45c+42

We move all terms to the left:

0-(3c^2-45c+42)=0

We add all the numbers together, and all the variables

-(3c^2-45c+42)=0

We get rid of parentheses

-3c^2+45c-42=0

a = -3; b = 45; c = -42;
Δ = b2-4ac
Δ = 452-4·(-3)·(-42)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-39}{2*-3}=\frac{-84}{-6}
=+14
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+39}{2*-3}=\frac{-6}{-6}
=1
$