(5r)^2-(4)^2=0

Solution for (5r)^2-(4)^2=0 equation:

(5r)^2-(4)^2=0

We add all the numbers together, and all the variables

5r^2-16=0

a = 5; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·5·(-16)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*5}=\frac{0-8\sqrt{5}}{10}
=-\frac{8\sqrt{5}}{10}
=-\frac{4\sqrt{5}}{5}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*5}=\frac{0+8\sqrt{5}}{10}
=\frac{8\sqrt{5}}{10}
=\frac{4\sqrt{5}}{5}
$