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.
$$2(3x^2+x-5)(2x^2-7x+3)=0.$$
 A: $2(3x^2 + x - 5)(2x^2 - 7x + 3) = 0$ means that either
$$3x^2 + x - 5 = 0 \hspace{1cm} \mathrm{or} \hspace{1cm} 2x^2 - 7x + 3 = 0$$
as obviously $2 \neq 0$. 
You can then solve both these equations using the quadratic formula or completing the square or however you want.
A: Hint: What you have written says these three numbers multiply to zero: $2(A)(B)=0$, where I've abbreviated the two factors you wrote into $A$ and $B$. This product can be zero only when either $A$ or $B$ are zero. So, your task is to find all the $x$'s which make $A$ and $B$ zero.
There are lots of ways to solve these two quadratic equations like factoring or using the quadratic formula. Definitely try factoring first, ( or maybe try both) and check that they are really zeros of the quadratic.
In any case, the discriminant of both of the quadratics is positive, indicating that there will be four real roots (and no complex roots).
A: Factor the polynomial $2x^2-7x+3$ by ruffini:
$$2(x-3) (2 x-1) (3 x^2+x-5)=0$$
So now you you have $x=3, x=1/2$ besides the two roots of $ (3 x^2+x-5)=0$ which are $x = 1/6 (-1-\sqrt{61})$ and $x=1/6 (-1+\sqrt{61})$.
If you don't know ruffini of course you can use directly the formula for the roots of a second degree equation in both polynomials. 
There are totally four solutions for $x$.
A: $$2(3x^2+x-5)(2x^2-7x+3)=0$$
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 $$x=\frac{-1+\sqrt {61}}{6},\frac{-1-\sqrt {61}}{6},3,\frac12$$
