1.9x^2+20x-500=0

Solution for 1.9x^2+20x-500=0 equation:

1.9x^2+20x-500=0

a = 1.9; b = 20; c = -500;
Δ = b2-4ac
Δ = 202-4·1.9·(-500)
Δ = 4200
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:

$\sqrt{\Delta}=\sqrt{4200}=\sqrt{100*42}=\sqrt{100}*\sqrt{42}=10\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{42}}{2*1.9}=\frac{-20-10\sqrt{42}}{3.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{42}}{2*1.9}=\frac{-20+10\sqrt{42}}{3.8}
$