\(a;x^2-4x+4=0\\ \left(x-2\right)^2=0\\ x-2=0\\ x=2\\ b;x^2-2\sqrt{3}x-4=0\\ \Delta=b^2-4ac=\left(-2\sqrt{3}\right)^2-4\cdot1\cdot\left(-4\right)=28>0\\ x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-2\sqrt{3}\right)-\sqrt{28}}{2\cdot1}=\sqrt{3}-\sqrt{7}\\ x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-2\sqrt{3}\right)+\sqrt{28}}{2\cdot1}=\sqrt{3}+\sqrt{7}\)
vậy phương trình có 2 nghiệm là: \(x_1=\sqrt{3}-\sqrt{7};x_2=\sqrt{3}+\sqrt{7}\)
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