10x^2+200x-500=0

Solution for 10x^2+200x-500=0 equation:

10x^2+200x-500=0

a = 10; b = 200; c = -500;
Δ = b2-4ac
Δ = 2002-4·10·(-500)
Δ = 60000
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:

$\sqrt{\Delta}=\sqrt{60000}=\sqrt{10000*6}=\sqrt{10000}*\sqrt{6}=100\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-100\sqrt{6}}{2*10}=\frac{-200-100\sqrt{6}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+100\sqrt{6}}{2*10}=\frac{-200+100\sqrt{6}}{20}
$