Like every quadratic equation
[tex]ax^2+bx+c=0[/tex]
where [tex]c=0[/tex]
[tex]ax^2+bx=0[/tex]
we can factor out [tex]x[/tex]
[tex]x(ax+b)=0[/tex]
And find the two solutions
[tex]x=0,\quad x=-\dfrac{b}{a}[/tex]
In this case, the solutions are
[tex]x=0,\quad x=\dfrac{3}{2}[/tex]
If you know that [tex]x_1,\ x_2[/tex] are the solutions to the quadratic equation
[tex]ax^2+bx+c=0[/tex]
then you can factor the poynomial as
[tex]a(x-x_1)(x-x_2)[/tex]
So, in this case, you have
[tex]10x^2-15x = 10(x-0)\left(x-\dfrac{3}{2}\right)=10x\left(x-\dfrac{3}{2}\right)[/tex]