(r-3)(7r-8)=0

Solution for (r-3)(7r-8)=0 equation:

(r-3)(7r-8)=0

We multiply parentheses ..

(+7r^2-8r-21r+24)=0

We get rid of parentheses

7r^2-8r-21r+24=0

We add all the numbers together, and all the variables

7r^2-29r+24=0

a = 7; b = -29; c = +24;
Δ = b2-4ac
Δ = -292-4·7·24
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-13}{2*7}=\frac{16}{14}
=1+1/7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+13}{2*7}=\frac{42}{14}
=3
$