Найти 0 функции у = log3(х2+5)−2 у = 4х - 3×2х – 4 у = √х2−4х+3

Решите задачу:

$1) \log_3(x^2+5)-2=0\\\\\log_3(x^2+5)=2\\\\3^2=x^2+5\\\\9=x^2+5\\\\x^2-4=0\\\\(x+2)(x-2)=0\\\\x_1=-2\\\\x_2=2\\\\2)4x-3\cdot 2x-4=0\\\\4x-6x-4=0\\\\-2x-4=0\\\\-2x=4\\\\x=-2$

$\displaystyle 3) \sqrt{x^2-4x+3}=0\\\\x^2-4x+3=0\\\\x_{1,2}= \frac{4\pm \sqrt{16-12} }{2}= \frac{4\pm 2}{2} =3,1$