x^+3x=(x+3)(x+1)

Solution for x^+3x=(x+3)(x+1) equation:

x^+3x=(x+3)(x+1)

We move all terms to the left:

x^+3x-((x+3)(x+1))=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

(+x^2+x+3x+3)

We get rid of parentheses

x^2+x+3x+3

We add all the numbers together, and all the variables

x^2+4x+3

Back to the equation:

-(x^2+4x+3)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-3=0

a = -1; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·(-1)·(-3)
Δ = -12
Delta is less than zero, so there is no solution for the equation