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

Solution for (r-2)(4r-3)=0 equation:

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

We multiply parentheses ..

(+4r^2-3r-8r+6)=0

We get rid of parentheses

4r^2-3r-8r+6=0

We add all the numbers together, and all the variables

4r^2-11r+6=0

a = 4; b = -11; c = +6;
Δ = b2-4ac
Δ = -112-4·4·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*4}=\frac{6}{8}
=3/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*4}=\frac{16}{8}
=2
$