(3r)^2-45=0

Solution for (3r)^2-45=0 equation:

(3r)^2-45=0

a = 3; b = 0; c = -45;
Δ = b2-4ac
Δ = 02-4·3·(-45)
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*3}=\frac{0-6\sqrt{15}}{6}
=-\frac{6\sqrt{15}}{6}
=-\sqrt{15}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*3}=\frac{0+6\sqrt{15}}{6}
=\frac{6\sqrt{15}}{6}
=\sqrt{15}
$