3x2+x-1900=0

Solution for 3x2+x-1900=0 equation:

3x^2+x-1900=0

a = 3; b = 1; c = -1900;
Δ = b2-4ac
Δ = 12-4·3·(-1900)
Δ = 22801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{22801}=151$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-151}{2*3}=\frac{-152}{6}
=-25+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+151}{2*3}=\frac{150}{6}
=25
$