3x^2+160x+2000=0

Solution for 3x^2+160x+2000=0 equation:

3x^2+160x+2000=0

a = 3; b = 160; c = +2000;
Δ = b2-4ac
Δ = 1602-4·3·2000
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-40}{2*3}=\frac{-200}{6}
=-33+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+40}{2*3}=\frac{-120}{6}
=-20
$