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0=(x+2)(x+1)
We move all terms to the left:
0-((x+2)(x+1))=0
We add all the numbers together, and all the variables
-((x+2)(x+1))=0
We multiply parentheses ..
-((+x^2+x+2x+2))=0
We calculate terms in parentheses: -((+x^2+x+2x+2)), so:We get rid of parentheses
(+x^2+x+2x+2)
We get rid of parentheses
x^2+x+2x+2
We add all the numbers together, and all the variables
x^2+3x+2
Back to the equation:
-(x^2+3x+2)
-x^2-3x-2=0
We add all the numbers together, and all the variables
-1x^2-3x-2=0
a = -1; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·(-1)·(-2)
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*-1}=\frac{2}{-2} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*-1}=\frac{4}{-2} =-2 $
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