Ответ:
Квадрат теңдеулерде \(z^2=a\) болса, \(a>0\) кезінде \(z=\pm\sqrt a\).
\(x^2=64\Rightarrow x=\pm8\).
\(y^2=0{,}09\Rightarrow y=\pm0{,}3\).
\(3x^2=48\Rightarrow x^2=16\Rightarrow x=\pm4\).
\(x^2=3\Rightarrow x=\pm\sqrt3\).
\(y^2-10=39\Rightarrow y^2=49\Rightarrow y=\pm7\).
\(x^2+5=30\Rightarrow x^2=25\Rightarrow x=\pm5\).
\(5t^2-3=77\Rightarrow t^2=16\Rightarrow t=\pm4\).
\(\frac12x^2=\frac89\Rightarrow x^2=\frac{16}{9}\Rightarrow x=\pm\frac43\).
