(x^2+3x)=18

Solution for (x^2+3x)=18 equation:

(x^2+3x)=18

We move all terms to the left:

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

We get rid of parentheses

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