(3x+36)(x-11)=(2x+34)(x-12)

Solution for (3x+36)(x-11)=(2x+34)(x-12) equation:

(3x+36)(x-11)=(2x+34)(x-12)

We move all terms to the left:

(3x+36)(x-11)-((2x+34)(x-12))=0

We multiply parentheses ..

(+3x^2-33x+36x-396)-((2x+34)(x-12))=0


We calculate terms in parentheses: -((2x+34)(x-12)), so:

(2x+34)(x-12)

We multiply parentheses ..

(+2x^2-24x+34x-408)

We get rid of parentheses

2x^2-24x+34x-408

We add all the numbers together, and all the variables

2x^2+10x-408

Back to the equation:

-(2x^2+10x-408)


We get rid of parentheses

3x^2-2x^2-33x+36x-10x-396+408=0

We add all the numbers together, and all the variables

x^2-7x+12=0

a = 1; b = -7; c = +12;
Δ = b2-4ac
Δ = -72-4·1·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*1}=\frac{6}{2}
=3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*1}=\frac{8}{2}
=4
$