96x^2-20x+1=0

Solution for 96x^2-20x+1=0 equation:

96x^2-20x+1=0

a = 96; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·96·1
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*96}=\frac{16}{192}
=1/12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*96}=\frac{24}{192}
=1/8
$