x=2,-6
x^2+4x-12
(x^2-2x)(+6x-12)
x(x-2)+6(x-2)
(x-2)(x+6)=0
I think you are talking about the x-intercepts. You can find the zeros of the equation of the parabola y=ax2 +bx+c by setting y equal to 0 and finding the corresponding x values. These will be the "roots" of the parabola.
(6,0) and (0,2)
To find the x-intercepts of the parabola described by the equation (x^2 + 4x - 12 = 0), we can use the quadratic formula (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}). Here, (a = 1), (b = 4), and (c = -12). Calculating the discriminant (b^2 - 4ac) gives (16 + 48 = 64). Therefore, the x-intercepts are (x = \frac{-4 \pm 8}{2}), resulting in (x = 2) and (x = -6).
"y = 2x2 - 12x + 6" is a quadratic equation which describes a parabola whose vertex occurs at the point (3, -12) and which has a range of -12 → ∞. It intercepts the x-axis at the points (3 - √6, 0) and (3 + √6, 0).
12
-12 +/- sq root 104
-5 + (-4) - 12 - 20 - (-10) + 6 = -25
12 - 78 + 2397543 - 124 + 3 = 2397356
18 + 12 + 0 - 4 - 10 = 16
8 plus 4 minus 12 divided by 1 is 0.
10 - 5 + 14324 + 12 + 30 - 1000 = 13371
y2-12=5x is an equation. When graphed, it is a parabola.