4r(3r-4)=0

Solution for 4r(3r-4)=0 equation:

4r(3r-4)=0

We multiply parentheses

12r^2-16r=0

a = 12; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·12·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*12}=\frac{0}{24}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*12}=\frac{32}{24}
=1+1/3
$