3(x+1)(x+2)=x-(2x-3)(x+4)

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

3(x+1)(x+2)=x-(2x-3)(x+4)

We move all terms to the left:

3(x+1)(x+2)-(x-(2x-3)(x+4))=0

We multiply parentheses ..

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


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

x-(2x-3)(x+4)

We multiply parentheses ..

-(+2x^2+8x-3x-12)+x

We get rid of parentheses

-2x^2-8x+3x+x+12

We add all the numbers together, and all the variables

-2x^2-4x+12

Back to the equation:

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


We multiply parentheses

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

We get rid of parentheses

3x^2+2x^2+4x+6x+3x-12+6=0

We add all the numbers together, and all the variables

5x^2+13x-6=0

a = 5; b = 13; c = -6;
Δ = b2-4ac
Δ = 132-4·5·(-6)
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*5}=\frac{-30}{10}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*5}=\frac{4}{10}
=2/5
$