10x^2-20x-99=0

Solution for 10x^2-20x-99=0 equation:

10x^2-20x-99=0

a = 10; b = -20; c = -99;
Δ = b2-4ac
Δ = -202-4·10·(-99)
Δ = 4360
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:

$\sqrt{\Delta}=\sqrt{4360}=\sqrt{4*1090}=\sqrt{4}*\sqrt{1090}=2\sqrt{1090}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{1090}}{2*10}=\frac{20-2\sqrt{1090}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{1090}}{2*10}=\frac{20+2\sqrt{1090}}{20}
$