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

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

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

We multiply parentheses ..

(+10r^2-12r-35r+42)=0

We get rid of parentheses

10r^2-12r-35r+42=0

We add all the numbers together, and all the variables

10r^2-47r+42=0

a = 10; b = -47; c = +42;
Δ = b2-4ac
Δ = -472-4·10·42
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-23}{2*10}=\frac{24}{20}
=1+1/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+23}{2*10}=\frac{70}{20}
=3+1/2
$