Question

  (√7-2)x*2-4x+(2+√7)=0

Answer 1

$(sqrt{7}-2)x^2-4x+(2+sqrt{7})=0\ (sqrt{7}-2)x^2-4x+(sqrt{7}+2)=0\ D=(-4)^2-4*(sqrt{7}-2)*(sqrt{7}+2)=16-4(sqrt{7}^2-2^2)=\ =16-4(7-4)=16-4*3=16-12=4\ sqrt{D} = sqrt{4} = 2\ x1=frac{4+2}{sqrt{7}-2} = frac{6}{sqrt{7}-2}\ x2 = frac{4-2}{sqrt{7}-2} = frac{2}{sqrt{7}-2}$