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4x^2+4x+1=x2
We move all terms to the left:
4x^2+4x+1-(x2)=0
We add all the numbers together, and all the variables
3x^2+4x+1=0
a = 3; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·3·1
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*3}=\frac{-6}{6} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*3}=\frac{-2}{6} =-1/3 $
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