How to find $x$ of this equation? I'm doing some equations and I'm not sure how to find $x$. Any idea? Step by step answers please, it's like my first time.
$$2(3x^2+x-5)(2x^2-7x+3)=0.$$

$$2(3x^2+x-5)(2x^2-7x+3)=0$$
so either $(3x^2+x-5)=0$ or $(2x^2-7x+3)=0$
so take it one by one.
part1
formula is for: $(ax^2+bx-c)=0$ ,$x=\dfrac{-b+\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$ ,
$\dfrac{-b-\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$
$$(3x^2+x-5)=0$$
$$x=\frac{-1+\sqrt{1^2-4\cdot3\cdot(-5)}}{2\cdot3}$$
$$x=\frac{-1+\sqrt {1+60}}{6}$$
$$x=\frac{-1+\sqrt {61}}{6}$$
and also
$$x=\frac{-1-\sqrt {61}}{6}$$
part2
$$(2x^2-7x+3)=0$$
from different method
$$(2x^2-6x-x+3)=0$$
$$2x(x-3)-1(x-3)=0$$
$$(x-3)(2x-1)=0$$
$$(x-3)=0 ,(2x-1)=0$$
so $x=3,x=\frac12$
so this euqtion have 4 roots $$x=\frac{-1+\sqrt {61}}{6},\frac{-1-\sqrt {61}}{6},3,\frac12$$
