Here is our quadratic:
[tex]3x^2+4x=2[/tex]
There are three different methods we can use. I'll explain each in depth.
Factoring
If you want to factor, bring everything over to the left side like so:
3x² + 4x = 2 | subtract 2 from each side
3x² + 4x - 2 = 0
Now, we need to find two numbers that add to our x coefficient 4 and multiply to make -6. (our x² coefficient times our constant) There's no whole number answer to that, so factoring isn't going to work.
If we had something like 2x² + 13x + 15 = 0, here's how we'd factor:
We want two numbers that add to 13 and multiply to 30...these would be 10 and 3.
Now we "split the middle" into these numbers.
2x² + 10x + 3x + 15 = 0
Factor the first two and last two terms. You should get the same thing inside the parentheses of each.
2x(x+5) + 3(x+5) = 0
(2x+3)(x+5) = 0
Any value of x which causes a factor to equal zero would be a solution.
2x + 3 = 0 ⇒ 2x = -3 ⇒ x = -3/2
x + 5 = 0 ⇒ x = -5
Completing the square
For this one, keep the constant on the right side.
3x² + 4x = 2
Our x² coefficient needs to be 1, so let's divide by 3.
x² + 4/3x = 2/3
Now we take our x coefficient, halve it, square it, and add it to each side.
Half of 4/3 is 2/3, square that to get 4/9...
x² + 4/3x + 4/9 = 2/3 + 4/9
x² + 4/3x + 4/9 = 10/9
Factor the perfect square trinomial on the left.
(Because we set this up as a perfect square trinomial, just halve the x coefficient and add that to x. Square it for your factored form.)
(x+2/3)² = 10/9 | Take the square root of each side
x+2/3 = ±√(10/9)
x+2/3 = ±√(10)/3 | Subtract 2/3 from each side.
x = ±√(10)/3-2/3
Under one denominator...
x = (-2±√10)/3 (answer C)
Using the quadratic formula
The solution to any quadratic [tex]ax^2+bx+c=0[/tex] is [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex].
In the case of our quadratic 3x² + 4x - 2 = 0...
a = 3, b = 4, and c = -2.
Plug these into our formula and simplify.
[tex]x=\frac{-4\pm\sqrt{4^2-4(3)(-2)}}{2(3)}=\frac{-4\pm\sqrt{16+24}}{6}=\frac{-4\pm\sqrt{40}}{6}=\frac{-4\pm2\sqrt{10}}{6}=\boxed{\frac{-2\pm\sqrt{10}}{3}}[/tex]
