0=12x^2-128x+240

Solution for 0=12x^2-128x+240 equation:

0=12x^2-128x+240

We move all terms to the left:

0-(12x^2-128x+240)=0

We add all the numbers together, and all the variables

-(12x^2-128x+240)=0

We get rid of parentheses

-12x^2+128x-240=0

a = -12; b = 128; c = -240;
Δ = b2-4ac
Δ = 1282-4·(-12)·(-240)
Δ = 4864
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:

$\sqrt{\Delta}=\sqrt{4864}=\sqrt{256*19}=\sqrt{256}*\sqrt{19}=16\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-16\sqrt{19}}{2*-12}=\frac{-128-16\sqrt{19}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+16\sqrt{19}}{2*-12}=\frac{-128+16\sqrt{19}}{-24}
$