3r2+1=4r

Solution for 3r2+1=4r equation:

3r^2+1=4r

We move all terms to the left:

3r^2+1-(4r)=0

a = 3; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·3·1
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2}{2*3}=\frac{2}{6}
=1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*3}=\frac{6}{6}
=1
$