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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

4x-((x+1)(x+2))+12=0

We multiply parentheses ..

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


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

(+x^2+2x+x+2)

We get rid of parentheses

x^2+2x+x+2

We add all the numbers together, and all the variables

x^2+3x+2

Back to the equation:

-(x^2+3x+2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+x+10=0

a = -1; b = 1; c = +10;
Δ = b2-4ac
Δ = 12-4·(-1)·10
Δ = 41
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{-(1)-\sqrt{41}}{2*-1}=\frac{-1-\sqrt{41}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{41}}{2*-1}=\frac{-1+\sqrt{41}}{-2}
$