3x^2+100x-500=0

Solution for 3x^2+100x-500=0 equation:

3x^2+100x-500=0

a = 3; b = 100; c = -500;
Δ = b2-4ac
Δ = 1002-4·3·(-500)
Δ = 16000
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:

$\sqrt{\Delta}=\sqrt{16000}=\sqrt{1600*10}=\sqrt{1600}*\sqrt{10}=40\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-40\sqrt{10}}{2*3}=\frac{-100-40\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+40\sqrt{10}}{2*3}=\frac{-100+40\sqrt{10}}{6}
$