R=-2x(3x-9)

Solution for R=-2x(3x-9) equation:

=-2R(3R-9)

We move all terms to the left:

-(-2R(3R-9))=0


We calculate terms in parentheses: -(-2R(3R-9)), so:

-2R(3R-9)

We multiply parentheses

-6R^2+18R

Back to the equation:

-(-6R^2+18R)


We get rid of parentheses

6R^2-18R=0

a = 6; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·6·0
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*6}=\frac{0}{12}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*6}=\frac{36}{12}
=3
$