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

Solution for (x+1)(x=-3)=0 equation:

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

We move all terms to the left:

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


We calculate terms in parentheses: +(x+1)(x-(-3)), so:

x+1)(x-(-3)

We add all the numbers together, and all the variables

x+1)(x+3

Back to the equation:

+(x+1)(x+3)


We multiply parentheses ..

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

We get rid of parentheses

x^2+3x+x+3=0

We add all the numbers together, and all the variables

x^2+4x+3=0

a = 1; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·1·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*1}=\frac{-2}{2}
=-1
$