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x(2)+3x+2=(x+2)
We move all terms to the left:
x(2)+3x+2-((x+2))=0
We add all the numbers together, and all the variables
x^2+3x-((x+2))+2=0
We calculate terms in parentheses: -((x+2)), so:We get rid of parentheses
(x+2)
We get rid of parentheses
x+2
Back to the equation:
-(x+2)
x^2+3x-x-2+2=0
We add all the numbers together, and all the variables
x^2+2x=0
a = 1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·1·0
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*1}=\frac{-4}{2} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*1}=\frac{0}{2} =0 $
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