16r^2-16r+3=0

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

16r^2-16r+3=0

a = 16; b = -16; c = +3;
Δ = b2-4ac
Δ = -162-4·16·3
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*16}=\frac{8}{32}
=1/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*16}=\frac{24}{32}
=3/4
$