-10x^2+20x+10=0

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

-10x^2+20x+10=0

a = -10; b = 20; c = +10;
Δ = b2-4ac
Δ = 202-4·(-10)·10
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{2}}{2*-10}=\frac{-20-20\sqrt{2}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{2}}{2*-10}=\frac{-20+20\sqrt{2}}{-20}
$