9x+36=(x+12)(x-1)

Solution for 9x+36=(x+12)(x-1) equation:

9x+36=(x+12)(x-1)

We move all terms to the left:

9x+36-((x+12)(x-1))=0

We multiply parentheses ..

-((+x^2-1x+12x-12))+9x+36=0


We calculate terms in parentheses: -((+x^2-1x+12x-12)), so:

(+x^2-1x+12x-12)

We get rid of parentheses

x^2-1x+12x-12

We add all the numbers together, and all the variables

x^2+11x-12

Back to the equation:

-(x^2+11x-12)


We add all the numbers together, and all the variables

9x-(x^2+11x-12)+36=0

We get rid of parentheses

-x^2+9x-11x+12+36=0

We add all the numbers together, and all the variables

-1x^2-2x+48=0

a = -1; b = -2; c = +48;
Δ = b2-4ac
Δ = -22-4·(-1)·48
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*-1}=\frac{16}{-2}
=-8
$