x+1=x^2-5

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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+5+1=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 $

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