3x^2+40x-1900=0

Solution for 3x^2+40x-1900=0 equation:

3x^2+40x-1900=0

a = 3; b = 40; c = -1900;
Δ = b2-4ac
Δ = 402-4·3·(-1900)
Δ = 24400
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:

$\sqrt{\Delta}=\sqrt{24400}=\sqrt{400*61}=\sqrt{400}*\sqrt{61}=20\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{61}}{2*3}=\frac{-40-20\sqrt{61}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{61}}{2*3}=\frac{-40+20\sqrt{61}}{6}
$