(x+3)(x+1)=1x+7

Solution for (x+3)(x+1)=1x+7 equation:

(x+3)(x+1)=1x+7

We move all terms to the left:

(x+3)(x+1)-(1x+7)=0

We add all the numbers together, and all the variables

(x+3)(x+1)-(x+7)=0

We get rid of parentheses

(x+3)(x+1)-x-7=0

We multiply parentheses ..

(+x^2+x+3x+3)-x-7=0

We add all the numbers together, and all the variables

(+x^2+x+3x+3)-1x-7=0

We get rid of parentheses

x^2+x+3x-1x+3-7=0

We add all the numbers together, and all the variables

x^2+3x-4=0

a = 1; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·1·(-4)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*1}=\frac{2}{2}
=1
$