Lời giải:
a)
\(3x^2+7x=0\)
\(\Leftrightarrow x(3x+7)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ 3x+7=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=0\\ x=-\frac{7}{3}\end{matrix}\right.\)
b)
\(-12x^2+15x=0\)
\(\Leftrightarrow 3x(-4x+5)=0\)
\(\Rightarrow \left[\begin{matrix} 3x=0\\ -4x+5=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=0\\ x=\frac{5}{4}\end{matrix}\right.\)