3r^2+1+4-16r=0

Solution for 3r^2+1+4-16r=0 equation:

3r^2+1+4-16r=0

We add all the numbers together, and all the variables

3r^2-16r+5=0

a = 3; b = -16; c = +5;
Δ = b2-4ac
Δ = -162-4·3·5
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-14}{2*3}=\frac{2}{6}
=1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+14}{2*3}=\frac{30}{6}
=5
$