MATH! THIS IS DUE SOON! PLEASE HELP ME NOW!?

f(x) = (1/3)x^2 + 4x + 10

Let's see if there are any real roots.

Discriminant: 4^2 - 4(1/3)(10) = 16 - (40/3) = 8/3. Two real roots. Now for the rest of the quadratic formula:

(-4 +/- sqrt(8/3)) / (2/3)

(-4 +/- (sqrt(8) / sqrt(3))) / (2/3)

(-4 +/- (sqrt(24) / 3))) / (2/3)

(-4 +/- (2 sqrt(6) / 3)) / (2/3)

(3/2)(-4 +/- (2 sqrt(6) / 3))

-6 +/- sqrt(6) (the coefficient of the right term is canceled out: 3/2 * 2/3 = 1)

There you go. x = -6 + sqrt(6), x = -6 - sqrt(6). Check box A) then type these values in.

0 = (1/3)(x^2 +12x +30)
0 = (x +6)^2 -6 . . . multiply by 3, rewrite in vertex form
x = -6 ± √6 . . . add 6, square root, subtract 6

x = {-6-√6, -6+√6}

x = [ - 4 ± √ (16 - 40 / 3 ) ] / (2/3) = [ - 4 ± 2 √( 2 / 3 ) ] / ( 2 / 3 ) = 3 [ - 2 ± √ ( 2 / 3 ) ]