9x+27=(x^2-6x)+(2x-3)

Solution for 9x+27=(x^2-6x)+(2x-3) equation:

9x+27=(x^2-6x)+(2x-3)

We move all terms to the left:

9x+27-((x^2-6x)+(2x-3))=0


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

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

We get rid of parentheses

x^2-6x+2x-3

We add all the numbers together, and all the variables

x^2-4x-3

Back to the equation:

-(x^2-4x-3)


We get rid of parentheses

-x^2+9x+4x+3+27=0

We add all the numbers together, and all the variables

-1x^2+13x+30=0

a = -1; b = 13; c = +30;
Δ = b2-4ac
Δ = 132-4·(-1)·30
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*-1}=\frac{-30}{-2}
=+15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*-1}=\frac{4}{-2}
=-2
$