-3r(2-r)=15

Solution for -3r(2-r)=15 equation:

-3r(2-r)=15

We move all terms to the left:

-3r(2-r)-(15)=0

We add all the numbers together, and all the variables

-3r(-1r+2)-15=0

We multiply parentheses

3r^2-6r-15=0

a = 3; b = -6; c = -15;
Δ = b2-4ac
Δ = -62-4·3·(-15)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{6}}{2*3}=\frac{6-6\sqrt{6}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{6}}{2*3}=\frac{6+6\sqrt{6}}{6}
$