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4x^2-4=4x
We move all terms to the left:
4x^2-4-(4x)=0
a = 4; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·4·(-4)
Δ = 80
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{5}}{2*4}=\frac{4-4\sqrt{5}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{5}}{2*4}=\frac{4+4\sqrt{5}}{8} $
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