18r-3(2r-5)=6(3r-9)9r

Solution for 18r-3(2r-5)=6(3r-9)9r equation:

18r-3(2r-5)=6(3r-9)9r

We move all terms to the left:

18r-3(2r-5)-(6(3r-9)9r)=0

We multiply parentheses

18r-6r-(6(3r-9)9r)+15=0


We calculate terms in parentheses: -(6(3r-9)9r), so:

6(3r-9)9r

We multiply parentheses

162r^2-486r

Back to the equation:

-(162r^2-486r)


We add all the numbers together, and all the variables

12r-(162r^2-486r)+15=0

We get rid of parentheses

-162r^2+12r+486r+15=0

We add all the numbers together, and all the variables

-162r^2+498r+15=0

a = -162; b = 498; c = +15;
Δ = b2-4ac
Δ = 4982-4·(-162)·15
Δ = 257724
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:

$\sqrt{\Delta}=\sqrt{257724}=\sqrt{36*7159}=\sqrt{36}*\sqrt{7159}=6\sqrt{7159}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(498)-6\sqrt{7159}}{2*-162}=\frac{-498-6\sqrt{7159}}{-324}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(498)+6\sqrt{7159}}{2*-162}=\frac{-498+6\sqrt{7159}}{-324}
$