x^2-16x+0.2=0

Solution for x^2-16x+0.2=0 equation:

x^2-16x+0.2=0

a = 1; b = -16; c = +0.2;
Δ = b2-4ac
Δ = -162-4·1·0.2
Δ = 255.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-\sqrt{255.2}}{2*1}=\frac{16-\sqrt{255.2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+\sqrt{255.2}}{2*1}=\frac{16+\sqrt{255.2}}{2}
$