(r-7)(r+5)=0

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

(r-7)(r+5)=0

We multiply parentheses ..

(+r^2+5r-7r-35)=0

We get rid of parentheses

r^2+5r-7r-35=0

We add all the numbers together, and all the variables

r^2-2r-35=0

a = 1; b = -2; c = -35;
Δ = b2-4ac
Δ = -22-4·1·(-35)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-12}{2*1}=\frac{-10}{2}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+12}{2*1}=\frac{14}{2}
=7
$