(7r-4)(8r-7)=0

Solution for (7r-4)(8r-7)=0 equation:

(7r-4)(8r-7)=0

We multiply parentheses ..

(+56r^2-49r-32r+28)=0

We get rid of parentheses

56r^2-49r-32r+28=0

We add all the numbers together, and all the variables

56r^2-81r+28=0

a = 56; b = -81; c = +28;
Δ = b2-4ac
Δ = -812-4·56·28
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-81)-17}{2*56}=\frac{64}{112}
=4/7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-81)+17}{2*56}=\frac{98}{112}
=7/8
$