3x^2+20x-5600=0

Solution for 3x^2+20x-5600=0 equation:

3x^2+20x-5600=0

a = 3; b = 20; c = -5600;
Δ = b2-4ac
Δ = 202-4·3·(-5600)
Δ = 67600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{67600}=260$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-260}{2*3}=\frac{-280}{6}
=-46+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+260}{2*3}=\frac{240}{6}
=40
$