r^2+3r=+9r+27

Solution for r^2+3r=+9r+27 equation:

r^2+3r=+9r+27

We move all terms to the left:

r^2+3r-(+9r+27)=0

We add all the numbers together, and all the variables

r^2+3r-(9r+27)=0

We get rid of parentheses

r^2+3r-9r-27=0

We add all the numbers together, and all the variables

r^2-6r-27=0

a = 1; b = -6; c = -27;
Δ = b2-4ac
Δ = -62-4·1·(-27)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12}{2*1}=\frac{-6}{2}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12}{2*1}=\frac{18}{2}
=9
$