32x^2-26x+5=0

Solution for 32x^2-26x+5=0 equation:

32x^2-26x+5=0

a = 32; b = -26; c = +5;
Δ = b2-4ac
Δ = -262-4·32·5
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-6}{2*32}=\frac{20}{64}
=5/16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+6}{2*32}=\frac{32}{64}
=1/2
$