(3r-4)(r-1)=0

Solution for (3r-4)(r-1)=0 equation:

(3r-4)(r-1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3r^2-7r+4=0

a = 3; b = -7; c = +4;
Δ = b2-4ac
Δ = -72-4·3·4
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*3}=\frac{6}{6}
=1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*3}=\frac{8}{6}
=1+1/3
$