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2x^2-4x+6=x^2+2x+1
We move all terms to the left:
2x^2-4x+6-(x^2+2x+1)=0
We get rid of parentheses
2x^2-x^2-4x-2x-1+6=0
We add all the numbers together, and all the variables
x^2-6x+5=0
a = 1; b = -6; c = +5;
Δ = b2-4ac
Δ = -62-4·1·5
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4}{2*1}=\frac{2}{2} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4}{2*1}=\frac{10}{2} =5 $
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