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

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



(x+1)(x+1)-3(x+1)=0
We multiply parentheses
(x+1)(x+1)-3x-3=0
We multiply parentheses ..
(+x^2+x+x+1)-3x-3=0
We get rid of parentheses
x^2+x+x-3x+1-3=0
We add all the numbers together, and all the variables
x^2-1x-2=0
a = 1; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·1·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*1}=\frac{-2}{2} =-1 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*1}=\frac{4}{2} =2 $

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