(r-8)(6r-5)=0

Solution for (r-8)(6r-5)=0 equation:

(r-8)(6r-5)=0

We multiply parentheses ..

(+6r^2-5r-48r+40)=0

We get rid of parentheses

6r^2-5r-48r+40=0

We add all the numbers together, and all the variables

6r^2-53r+40=0

a = 6; b = -53; c = +40;
Δ = b2-4ac
Δ = -532-4·6·40
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-43}{2*6}=\frac{10}{12}
=5/6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+43}{2*6}=\frac{96}{12}
=8
$