x(x+1)+(x=2)+(x+3)=-26

Solution for x(x+1)+(x=2)+(x+3)=-26 equation:

x(x+1)+(x=2)+(x+3)=-26

We move all terms to the left:

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

We multiply parentheses

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


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

x-2+(x+3)

determiningTheFunctionDomain
x+(x+3)-2

We get rid of parentheses

x+x+3-2

We add all the numbers together, and all the variables

2x+1

Back to the equation:

+(2x+1)


We get rid of parentheses

x^2+x+2x+1=0

We add all the numbers together, and all the variables

x^2+3x+1=0

a = 1; b = 3; c = +1;
Δ = b2-4ac
Δ = 32-4·1·1
Δ = 5
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{-(3)-\sqrt{5}}{2*1}=\frac{-3-\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{5}}{2*1}=\frac{-3+\sqrt{5}}{2}
$