2x+36=(x+3)(x+4)

Solution for 2x+36=(x+3)(x+4) equation:

2x+36=(x+3)(x+4)

We move all terms to the left:

2x+36-((x+3)(x+4))=0

We multiply parentheses ..

-((+x^2+4x+3x+12))+2x+36=0


We calculate terms in parentheses: -((+x^2+4x+3x+12)), so:

(+x^2+4x+3x+12)

We get rid of parentheses

x^2+4x+3x+12

We add all the numbers together, and all the variables

x^2+7x+12

Back to the equation:

-(x^2+7x+12)


We add all the numbers together, and all the variables

2x-(x^2+7x+12)+36=0

We get rid of parentheses

-x^2+2x-7x-12+36=0

We add all the numbers together, and all the variables

-1x^2-5x+24=0

a = -1; b = -5; c = +24;
Δ = b2-4ac
Δ = -52-4·(-1)·24
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*-1}=\frac{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*-1}=\frac{16}{-2}
=-8
$