(2r-3)(6r-7)=0

Solution for (2r-3)(6r-7)=0 equation:

(2r-3)(6r-7)=0

We multiply parentheses ..

(+12r^2-14r-18r+21)=0

We get rid of parentheses

12r^2-14r-18r+21=0

We add all the numbers together, and all the variables

12r^2-32r+21=0

a = 12; b = -32; c = +21;
Δ = b2-4ac
Δ = -322-4·12·21
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4}{2*12}=\frac{28}{24}
=1+1/6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4}{2*12}=\frac{36}{24}
=1+1/2
$