x^2+3x=6x+18

Solution for x^2+3x=6x+18 equation:

x^2+3x=6x+18

We move all terms to the left:

x^2+3x-(6x+18)=0

We get rid of parentheses

x^2+3x-6x-18=0

We add all the numbers together, and all the variables

x^2-3x-18=0

a = 1; b = -3; c = -18;
Δ = b2-4ac
Δ = -32-4·1·(-18)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*1}=\frac{12}{2}
=6
$