(r-5)(r+5)=25

Solution for (r-5)(r+5)=25 equation:

(r-5)(r+5)=25

We move all terms to the left:

(r-5)(r+5)-(25)=0

We use the square of the difference formula

r^2-25-25=0

We add all the numbers together, and all the variables

r^2-50=0

a = 1; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·1·(-50)
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*1}=\frac{0-10\sqrt{2}}{2}
=-\frac{10\sqrt{2}}{2}
=-5\sqrt{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*1}=\frac{0+10\sqrt{2}}{2}
=\frac{10\sqrt{2}}{2}
=5\sqrt{2}
$