0=12x^2-48x+42

Solution for 0=12x^2-48x+42 equation:

0=12x^2-48x+42

We move all terms to the left:

0-(12x^2-48x+42)=0

We add all the numbers together, and all the variables

-(12x^2-48x+42)=0

We get rid of parentheses

-12x^2+48x-42=0

a = -12; b = 48; c = -42;
Δ = b2-4ac
Δ = 482-4·(-12)·(-42)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12\sqrt{2}}{2*-12}=\frac{-48-12\sqrt{2}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12\sqrt{2}}{2*-12}=\frac{-48+12\sqrt{2}}{-24}
$