Help finding x-intercepts

3x^2+6x-7

The answer has radicals and I'm confused. Can someone help?

3x^2 + 6x - 7 = 0
If it's not factorable, you can either complete the square or use the quadratic formula. Let's use the quadratic formula for this one.

x = (-b +/- sqrt(b^2 - 4ac)) / 2a
x = (-6 +/- sqrt(6^2 - 4(3)(-7)) / 2(3)
x = (-6 +/- sqrt(36 + 84)) / 6
x = (-6 +/- sqrt(120)) / 6
x = (-6 +/- 2 sqrt(30)) / 6
x = -1 +/- (sqrt(30) / 3)

So the x-intercepts are: (-1 + sqrt(30) / 3, 0) and (-1 - sqrt(30) / 3, 0).

If you wanted to complete the square to solve it instead, do the following:

3x^2 + 6x - 7 = 0
x^2 + 2x - 7/3 = 0
(x^2 + 2x + 1) = 1 + 7/3
(x + 1)^2 = 10/3
x + 1 = +/- sqrt(10/3)
x = -1 +/- sqrt(10/3)
x = -1 +/- sqrt(10) / sqrt(3)
x = -1 +/- (sqrt(10) * sqrt(3)) / 3
x = -1 +/- sqrt(30) / 3

If your line has the equation y = 3x² + 6x - 7 then to find the x intercepts you set y = 0.

So 0 = 3x² + 6x - 7

Like you said (or implied), it won't factorize so we should use the quadratic formula to solve it.

x = [-6 +/- sqrt(6² - 4(3 x -7))] / 2(3)

x = [-6 +/- sqrt(120)] / 6

x = [-6 +/- 2sqrt(30)] / 6

x = [-3 +/- sqrt(30)] / 3

x = -1 +/- sqrt(30)/3

So those are your two answers.

x = -1 + sqrt(30)/3 or
x = -1 - sqrt(30)/3