6r=3(r-4)r

Solution for 6r=3(r-4)r equation:

6r=3(r-4)r

We move all terms to the left:

6r-(3(r-4)r)=0


We calculate terms in parentheses: -(3(r-4)r), so:

3(r-4)r

We multiply parentheses

3r^2-12r

Back to the equation:

-(3r^2-12r)


We get rid of parentheses

-3r^2+6r+12r=0

We add all the numbers together, and all the variables

-3r^2+18r=0

a = -3; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·(-3)·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*-3}=\frac{-36}{-6}
=+6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*-3}=\frac{0}{-6}
=0
$