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