By the quadratic formula,
\(x = \frac{-5 \pm \sqrt{5^2 - 4(3)(-2)}}{2} = \frac{-5 \pm 7}{2}\)
so the roots are -6 and 1.
first, we can do 3x2.
so this would be 6+5x-2=0
Now, we can move all the numbers to the other side by adding or subtracting them on each side
6-6+5x-2+2=0-6+2
or
5x = -4
then divide both sides by 5
x = -4/5
To factor the left-hand side, we need 2 numbers that sum to 5 and two numbers that multiply to \(3 \times -2 = -6\). The only two numbers that work are -1 and 6.
Now, rewrite this equation to \(3x^2 - x + 6x - 2 = 0\). Just factor \(3x^2 - x \) to \(x(3x - 1)\) and factor \(6x - 2\) to \(2(3x - 1)\).
Adding both of these gives us \((x+2)(3x-1) = 0\).
If we want the product to be 0, x + 2 = 0, or 3x - 1 = 0.
Can you take it from here?