(x^2+9x+18)=(x+6)

Solution for (x^2+9x+18)=(x+6) equation:

(x^2+9x+18)=(x+6)

We move all terms to the left:

(x^2+9x+18)-((x+6))=0

We get rid of parentheses

x^2+9x-((x+6))+18=0


We calculate terms in parentheses: -((x+6)), so:

(x+6)

We get rid of parentheses

x+6

Back to the equation:

-(x+6)


We get rid of parentheses

x^2+9x-x-6+18=0

We add all the numbers together, and all the variables

x^2+8x+12=0

a = 1; b = 8; c = +12;
Δ = b2-4ac
Δ = 82-4·1·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*1}=\frac{-12}{2}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*1}=\frac{-4}{2}
=-2
$