25-4r^2=0

Solution for 25-4r^2=0 equation:

25-4r^2=0

a = -4; b = 0; c = +25;
Δ = b2-4ac
Δ = 02-4·(-4)·25
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*-4}=\frac{-20}{-8}
=2+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*-4}=\frac{20}{-8}
=-2+1/2
$