16x^2-16x+0.5=0

Solution for 16x^2-16x+0.5=0 equation:

16x^2-16x+0.5=0

a = 16; b = -16; c = +0.5;
Δ = b2-4ac
Δ = -162-4·16·0.5
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{14}}{2*16}=\frac{16-4\sqrt{14}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{14}}{2*16}=\frac{16+4\sqrt{14}}{32}
$