-x^2-9x+18=(x-3)(x+6)

Solution for -x^2-9x+18=(x-3)(x+6) equation:

-x^2-9x+18=(x-3)(x+6)

We move all terms to the left:

-x^2-9x+18-((x-3)(x+6))=0

We add all the numbers together, and all the variables

-1x^2-9x-((x-3)(x+6))+18=0

We multiply parentheses ..

-1x^2-((+x^2+6x-3x-18))-9x+18=0


We calculate terms in parentheses: -((+x^2+6x-3x-18)), so:

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

We get rid of parentheses

x^2+6x-3x-18

We add all the numbers together, and all the variables

x^2+3x-18

Back to the equation:

-(x^2+3x-18)


We add all the numbers together, and all the variables

-1x^2-9x-(x^2+3x-18)+18=0

We get rid of parentheses

-1x^2-x^2-9x-3x+18+18=0

We add all the numbers together, and all the variables

-2x^2-12x+36=0

a = -2; b = -12; c = +36;
Δ = b2-4ac
Δ = -122-4·(-2)·36
Δ = 432
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{3}}{2*-2}=\frac{12-12\sqrt{3}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{3}}{2*-2}=\frac{12+12\sqrt{3}}{-4}
$