25/36-y^2=0 решите пожалуйста

Решите задачу:

$\frac{25}{36}-y^2=0 \\ -y^2+ \frac{25}{36}=0 \\ -y^2=- \frac{25}{25} \\ y^2= \frac{25}{36} \\ y= \sqrt{ \frac{25}{36} } \\ y= \frac{ \sqrt{25} }{ \sqrt{36} } \\ |y|= \frac{5}{6}$

$\frac{25}{36}-y^2=0 \\ (\frac{5}{6})^2-y^2=0 \\ ( \frac{5}{6}-y)( \frac{5}{6}+y)=0 \\ \\ \frac{5}{6}-y=0 \\ -y=- \frac{5}{6} \\ y= \frac{5}{6} \\ \\ \frac{5}{6}+y=0 \\ y=- \frac{5}{6} \\ \\ |y|= \frac{5}{6}$