[(x^2)-20x+(23/3)]'-√576+4!=0

[(x^2)-20x+(23/3)]'-√576+4!=0

$$\\\left(x^2-20x+\frac{23}{3}\right)-\sqrt{576}+4!=0\\\\
\left(x^2-20x+\frac{23}{3}\right)-24+24=0\\\\
x^2-20x+\frac{23}{3}=0\\\\
3x^2-60x+23=0\\\\
x=\frac{60\pm\sqrt{3600-276}}{6}\\\\
x=\frac{60\pm\sqrt{3324}}{6}\\\\
etc$$

 [(x^2)-20x+(23/3)]'-√576+4!=0?

What is the apostrophy for?

sorry must have put that there by accident...