(2r-6)(5r-7)=0

Solution for (2r-6)(5r-7)=0 equation:

(2r-6)(5r-7)=0

We multiply parentheses ..

(+10r^2-14r-30r+42)=0

We get rid of parentheses

10r^2-14r-30r+42=0

We add all the numbers together, and all the variables

10r^2-44r+42=0

a = 10; b = -44; c = +42;
Δ = b2-4ac
Δ = -442-4·10·42
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{-(-44)-16}{2*10}=\frac{28}{20}
=1+2/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+16}{2*10}=\frac{60}{20}
=3
$