-4r^2+81=0

Solution for -4r^2+81=0 equation:

-4r^2+81=0

a = -4; b = 0; c = +81;
Δ = b2-4ac
Δ = 02-4·(-4)·81
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36}{2*-4}=\frac{-36}{-8}
=4+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*-4}=\frac{36}{-8}
=-4+1/2
$