a) Ta có: \(6x^2-15x+\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow6x^2-15x+4x^2-25=0\)
\(\Leftrightarrow10x^2-15x-25=0\)
\(\Leftrightarrow10x^2-25x+10x-25=0\)
\(\Leftrightarrow5x\left(2x-5\right)+2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{2}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{5}{2};-\frac{2}{5}\right\}\)
b) Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
mà \(x^2+1\ne0\forall x\)
nên 2x+3=0
\(\Leftrightarrow2x=-3\)
hay \(x=-\frac{3}{2}\)
Vậy: \(x=-\frac{3}{2}\)