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

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

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

We add all the numbers together, and all the variables

(-1r+7)(-1r+6)=0

We multiply parentheses ..

(+r^2-6r-7r+42)=0

We get rid of parentheses

r^2-6r-7r+42=0

We add all the numbers together, and all the variables

r^2-13r+42=0

a = 1; b = -13; c = +42;
Δ = b2-4ac
Δ = -132-4·1·42
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-1}{2*1}=\frac{12}{2}
=6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+1}{2*1}=\frac{14}{2}
=7
$