0=3x^2+4x-2400

Solution for 0=3x^2+4x-2400 equation:

0=3x^2+4x-2400

We move all terms to the left:

0-(3x^2+4x-2400)=0

We add all the numbers together, and all the variables

-(3x^2+4x-2400)=0

We get rid of parentheses

-3x^2-4x+2400=0

a = -3; b = -4; c = +2400;
Δ = b2-4ac
Δ = -42-4·(-3)·2400
Δ = 28816
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:

$\sqrt{\Delta}=\sqrt{28816}=\sqrt{16*1801}=\sqrt{16}*\sqrt{1801}=4\sqrt{1801}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{1801}}{2*-3}=\frac{4-4\sqrt{1801}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{1801}}{2*-3}=\frac{4+4\sqrt{1801}}{-6}
$