2x^2-2x-0.5=0

Solution for 2x^2-2x-0.5=0 equation:

2x^2-2x-0.5=0

a = 2; b = -2; c = -0.5;
Δ = b2-4ac
Δ = -22-4·2·(-0.5)
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{2}}{2*2}=\frac{2-2\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{2}}{2*2}=\frac{2+2\sqrt{2}}{4}
$