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(x^2+2x+1)+(x+1)(3x-1)=0
We get rid of parentheses
x^2+2x+(x+1)(3x-1)+1=0
We multiply parentheses ..
x^2+(+3x^2-1x+3x-1)+2x+1=0
We get rid of parentheses
x^2+3x^2-1x+3x+2x-1+1=0
We add all the numbers together, and all the variables
4x^2+4x=0
a = 4; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·4·0
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*4}=\frac{-8}{8} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*4}=\frac{0}{8} =0 $
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