(2x+3)(4x+1)=(3x+12)(6x-9)

Solution for (2x+3)(4x+1)=(3x+12)(6x-9) equation:

(2x+3)(4x+1)=(3x+12)(6x-9)

We move all terms to the left:

(2x+3)(4x+1)-((3x+12)(6x-9))=0

We multiply parentheses ..

(+8x^2+2x+12x+3)-((3x+12)(6x-9))=0


We calculate terms in parentheses: -((3x+12)(6x-9)), so:

(3x+12)(6x-9)

We multiply parentheses ..

(+18x^2-27x+72x-108)

We get rid of parentheses

18x^2-27x+72x-108

We add all the numbers together, and all the variables

18x^2+45x-108

Back to the equation:

-(18x^2+45x-108)


We get rid of parentheses

8x^2-18x^2+2x+12x-45x+3+108=0

We add all the numbers together, and all the variables

-10x^2-31x+111=0

a = -10; b = -31; c = +111;
Δ = b2-4ac
Δ = -312-4·(-10)·111
Δ = 5401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-\sqrt{5401}}{2*-10}=\frac{31-\sqrt{5401}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{5401}}{2*-10}=\frac{31+\sqrt{5401}}{-20}
$