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(2x+3)(x+1)=(x+1)x
We move all terms to the left:
(2x+3)(x+1)-((x+1)x)=0
We multiply parentheses ..
(+2x^2+2x+3x+3)-((x+1)x)=0
We calculate terms in parentheses: -((x+1)x), so:We get rid of parentheses
(x+1)x
We multiply parentheses
x^2+x
Back to the equation:
-(x^2+x)
2x^2-x^2+2x+3x-x+3=0
We add all the numbers together, and all the variables
x^2+4x+3=0
a = 1; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·1·3
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*1}=\frac{-6}{2} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*1}=\frac{-2}{2} =-1 $
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