0=4(3x^2-26x+42)

Solution for 0=4(3x^2-26x+42) equation:

0=4(3x^2-26x+42)

We move all terms to the left:

0-(4(3x^2-26x+42))=0

We add all the numbers together, and all the variables

-(4(3x^2-26x+42))=0


We calculate terms in parentheses: -(4(3x^2-26x+42)), so:

4(3x^2-26x+42)

We multiply parentheses

12x^2-104x+168

Back to the equation:

-(12x^2-104x+168)


We get rid of parentheses

-12x^2+104x-168=0

a = -12; b = 104; c = -168;
Δ = b2-4ac
Δ = 1042-4·(-12)·(-168)
Δ = 2752
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:

$\sqrt{\Delta}=\sqrt{2752}=\sqrt{64*43}=\sqrt{64}*\sqrt{43}=8\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(104)-8\sqrt{43}}{2*-12}=\frac{-104-8\sqrt{43}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(104)+8\sqrt{43}}{2*-12}=\frac{-104+8\sqrt{43}}{-24}
$