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x2 + 4x - 12 = 0 so (x+6)*(x-2) = 0 so x = -6 or x = 2

To solve the quadratic equation x2 + 4x - 12 = 0, you can use two different methods:

1: Reduce it to its factors:

x2 + 4x - 12 = 0 is the same as (x + 6)(x - 2) = 0. But this can only be correct if one of the factors is zero which only happens when x = -6 or x = 2. That is the solution.

2: Use the quadratic solution formula:

x2 + 4x - 12 = 0 is of the form ax2 + bx + c = 0 with a = 1, b = 4, and c = -12.

Substitute these values into the formula

x = (-b ± sqrt(b2 - 4ac)) / 2a and you get

x = (-4 ± sqrt(16 - (-48)) / 2

= (-4 ±sqrt(64)) / 2

= (-4 + 8) / 2 or (-4 - 8) / 2

= 4 / 2 or -12 / 2

so x = 2 or x = -6

Q: X2 plus 4x-12 equals 0

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x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0

x2 + 49 = 0 ∴ x2 = -49 ∴ x = 7i

-2

No.

168

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x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.

x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0

no

x2 + 49 = 0 ∴ x2 = -49 ∴ x = 7i

-2

No.

168

x2 + 10x = 0 x2 + 10x + 25 = 25 (x + 5)2 = 25 x + 5 = +-5 x1 = 0 x2 =10

If: y = x2+20x+100 and x2-20x+100 Then: x2+20x+100 = x2-20x+100 So: 40x = 0 => x = 0 When x = 0 then y = 100 Therefore point of intersection: (0, 100)

x2-5-4x2+3x = 0 -3x2+3x-5 = 0 or as 3x2-3x+5 = 0

x = -3.27 or -6.73

Yes