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

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

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

We multiply parentheses ..

(+7r^2+35r-5r-25)=0

We get rid of parentheses

7r^2+35r-5r-25=0

We add all the numbers together, and all the variables

7r^2+30r-25=0

a = 7; b = 30; c = -25;
Δ = b2-4ac
Δ = 302-4·7·(-25)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-40}{2*7}=\frac{-70}{14}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+40}{2*7}=\frac{10}{14}
=5/7
$