(3r*1)*(r*2)=65

Solution for (3r*1)*(r*2)=65 equation:

(3r*1)(r*2)=65

We move all terms to the left:

(3r*1)(r*2)-(65)=0

We add all the numbers together, and all the variables

(+3r*1)(+r*2)-65=0

We multiply parentheses ..

(+6r^2)-65=0

We get rid of parentheses

6r^2-65=0

a = 6; b = 0; c = -65;
Δ = b2-4ac
Δ = 02-4·6·(-65)
Δ = 1560
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:

$\sqrt{\Delta}=\sqrt{1560}=\sqrt{4*390}=\sqrt{4}*\sqrt{390}=2\sqrt{390}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{390}}{2*6}=\frac{0-2\sqrt{390}}{12}
=-\frac{2\sqrt{390}}{12}
=-\frac{\sqrt{390}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{390}}{2*6}=\frac{0+2\sqrt{390}}{12}
=\frac{2\sqrt{390}}{12}
=\frac{\sqrt{390}}{6}
$