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

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

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

We use the square of the difference formula

r^2-25=0

a = 1; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·1·(-25)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*1}=\frac{-10}{2}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*1}=\frac{10}{2}
=5
$