0=3x^2-26x+42

Solution for 0=3x^2-26x+42 equation:

0=3x^2-26x+42

We move all terms to the left:

0-(3x^2-26x+42)=0

We add all the numbers together, and all the variables

-(3x^2-26x+42)=0

We get rid of parentheses

-3x^2+26x-42=0

a = -3; b = 26; c = -42;
Δ = b2-4ac
Δ = 262-4·(-3)·(-42)
Δ = 172
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{43}}{2*-3}=\frac{-26-2\sqrt{43}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{43}}{2*-3}=\frac{-26+2\sqrt{43}}{-6}
$