\(3x-15=2x^2-10x=>3x-15-2x^2+10x=0=>13x-2x^2-15=0=>...\)
\(3x-15=2x\left(x-5\right)\\ \Leftrightarrow3x-15=2x^2-10x\\ \Leftrightarrow3x+2x^2+10x=15\\ \Leftrightarrow13x+2x^2=15\\ \Leftrightarrow x\left(13+2x\right)=15\\ \Leftrightarrow\left[{}\begin{matrix}x=15\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=1\end{matrix}\right.\)
3x-15=2x(x-5)
\(3x-15=2x^2-10x\)
\(-2x^2+10x+3x-15=0\)
\(2x\left(-x+5\right)-3\left(-x+5\right)=0\)
\(\left(2x-3\right)\left(-x+5\right)=0\)
=> 2x-3=0 hoặc -x+5=0
2x=3 -x=-5
x=\(\dfrac{3}{2}\) x=5
Vậy ...
\(3x-15=2x(x-5)\\3x-15-2x(x-5)=0\\3(x-5)-2x(x-5)=0\\(x-5)(3-2x)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-3}{-2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)