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

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

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

We multiply parentheses ..

(+4r^2-3r-12r+9)=0

We get rid of parentheses

4r^2-3r-12r+9=0

We add all the numbers together, and all the variables

4r^2-15r+9=0

a = 4; b = -15; c = +9;
Δ = b2-4ac
Δ = -152-4·4·9
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-9}{2*4}=\frac{6}{8}
=3/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+9}{2*4}=\frac{24}{8}
=3
$