5x^2+100x-500=0

Solution for 5x^2+100x-500=0 equation:

5x^2+100x-500=0

a = 5; b = 100; c = -500;
Δ = b2-4ac
Δ = 1002-4·5·(-500)
Δ = 20000
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:

$\sqrt{\Delta}=\sqrt{20000}=\sqrt{10000*2}=\sqrt{10000}*\sqrt{2}=100\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-100\sqrt{2}}{2*5}=\frac{-100-100\sqrt{2}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+100\sqrt{2}}{2*5}=\frac{-100+100\sqrt{2}}{10}
$