3x^2+28x-2400=0

Solution for 3x^2+28x-2400=0 equation:

3x^2+28x-2400=0

a = 3; b = 28; c = -2400;
Δ = b2-4ac
Δ = 282-4·3·(-2400)
Δ = 29584
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{29584}=172$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-172}{2*3}=\frac{-200}{6}
=-33+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+172}{2*3}=\frac{144}{6}
=24
$