16x^2-257x+16=0

Solution for 16x^2-257x+16=0 equation:

16x^2-257x+16=0

a = 16; b = -257; c = +16;
Δ = b2-4ac
Δ = -2572-4·16·16
Δ = 65025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{65025}=255$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-257)-255}{2*16}=\frac{2}{32}
=1/16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-257)+255}{2*16}=\frac{512}{32}
=16
$