3x^2+20x-1800=0

Solution for 3x^2+20x-1800=0 equation:

3x^2+20x-1800=0

a = 3; b = 20; c = -1800;
Δ = b2-4ac
Δ = 202-4·3·(-1800)
Δ = 22000
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:

$\sqrt{\Delta}=\sqrt{22000}=\sqrt{400*55}=\sqrt{400}*\sqrt{55}=20\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{55}}{2*3}=\frac{-20-20\sqrt{55}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{55}}{2*3}=\frac{-20+20\sqrt{55}}{6}
$