(x+1)=x^2-5

Solution for (x+1)=x^2-5 equation:

(x+1)=x^2-5

We move all terms to the left:

(x+1)-(x^2-5)=0

We get rid of parentheses

-x^2+x+1+5=0

We add all the numbers together, and all the variables

-1x^2+x+6=0

a = -1; b = 1; c = +6;
Δ = b2-4ac
Δ = 12-4·(-1)·6
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{-(1)-5}{2*-1}=\frac{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*-1}=\frac{4}{-2}
=-2
$