what is the answer to 3x^2+2x+2=0
Equations of the form ax2+bx+c = 0 are known as quadratic equations.
If they have real roots then they can be factored knowing that
(x+r1)(x+r2) = 0 multiplied out gives you
x2+(r1+r2)x + (r1*r2) = 0
where a =1, b=r2 +r2, and c = r1*r2
To try to factor the equation you listed,
since a = 3 there would be two terms that would have to multiply together to equal 3, the only factors of a prime number are one and itself so 1 and 3 are your only choices.
and because c = 2 the only two choices are 1 and 2 in this case.
So, you have two possibilities:
(3x+2)(x+1) = 0 or (x+2)(3x+1) = 0
Now if you multiply either of those out you don't get your equation at the top because the middle term "b" adds up to 5 in one case and 7 in the other.
That indicates that you don't have any real solutions for this equation and that you need to use the quadratic formula to discover the answer.
In case you forgot it's: x = [-b ± √(b2-4ac)]/2a
I'll leave the calculation to you but if you plug in your values you should end up with:
x = -1/3 ± (√5)i/3
In factor form that would be:
(x+1/3+(√5)i/3)(x+1/3-(√5)i/3) = 0
Now for kicks let's multiply it out, I'm going to seperate them by lines for clarity but just know that I'm multiplying the first term in the first parentheses by all of the terms in the next, then the second, and so on:
x2 + (1/3)x - [(√5)i/3]x +
(1/3)x + 1/9 - (√5)i/9 +
[(√5)i/3]x + (√5)i/9 - [(√5)2i2]/9 = 0
It's important to recognize here that the numerator on the last term is the square root of 5 squared
(√5)2 = 5 times i2 = (√-1)2 = - 1 , that means the last term is equal to -5/9.
Notice the sign in front of it and it becomes 5/9.
So, to remove the terms that cancel you are left with:
x2 + (1/3)x + (1/3)x + 1/9 + 5/9 = 0 or
x2 + (2/3)x + (6/9) = 0
x2+(2/3)x + 2/3 = 0 now multiply through by 3 and you're back to where you started:
3x2 + 2x + 2 = 0
i know you probably wanted the short answer for this question but its important to understand how you can go both ways in order to check your answer and to get used to dealing with complex numbers in general.