0=3x^2+48x-3456

Solution for 0=3x^2+48x-3456 equation:

0=3x^2+48x-3456

We move all terms to the left:

0-(3x^2+48x-3456)=0

We add all the numbers together, and all the variables

-(3x^2+48x-3456)=0

We get rid of parentheses

-3x^2-48x+3456=0

a = -3; b = -48; c = +3456;
Δ = b2-4ac
Δ = -482-4·(-3)·3456
Δ = 43776
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:

$\sqrt{\Delta}=\sqrt{43776}=\sqrt{2304*19}=\sqrt{2304}*\sqrt{19}=48\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-48\sqrt{19}}{2*-3}=\frac{48-48\sqrt{19}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48\sqrt{19}}{2*-3}=\frac{48+48\sqrt{19}}{-6}
$