16x^2-20x+3=0

Solution for 16x^2-20x+3=0 equation:

16x^2-20x+3=0

a = 16; b = -20; c = +3;
Δ = b2-4ac
Δ = -202-4·16·3
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{13}}{2*16}=\frac{20-4\sqrt{13}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{13}}{2*16}=\frac{20+4\sqrt{13}}{32}
$