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 = 4000
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:

$\sqrt{\Delta}=\sqrt{4000}=\sqrt{400*10}=\sqrt{400}*\sqrt{10}=20\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-20\sqrt{10}}{2*3}=\frac{-100-20\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+20\sqrt{10}}{2*3}=\frac{-100+20\sqrt{10}}{6}
$