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Solve x² + 7x + 12 = 0

The solutions of x² + 7x + 12 = 0 are x = -3 or x = -4.

x² +x += 0
x2+7x+12=0⇒x=−3, −4x^2 + 7x + 12 = 0 \Rightarrow x = -3,\ -4
Discriminant D = 1 · x ≈ -3, -4

Change a, b or c (whole numbers, a not 0) and the exact solutions follow.

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Graph of y = x² + 7x + 12-6-5-4-3-2-12468x = -3x = -4vertexy = x² + 7x + 12
Graph of y = x² + 7x + 12. The curve crosses the x-axis at x = -3 and x = -4; the vertex is at (-3.5, -0.25).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 1, b = 7, c = 12.

    x2+7x+12=0x^2 + 7x + 12 = 0
  2. 2

    Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.

    D=72−4⋅1⋅12=49−48=1D = 7^2 - 4\cdot1\cdot12 = 49 - 48 = 1
  3. 3

    D is positive, so there are two different real solutions.

  4. 4

    Method 1, factoring. The roots are rational, so the left side factors:

    (x+3)(x+4)=0\left(x + 3\right)\left(x + 4\right) = 0
  5. 5

    A product is zero when one of its factors is zero, so set each factor equal to 0 and solve:

    x+3=0⇒x=−3x+4=0⇒x=−4x + 3 = 0 \Rightarrow x = -3\qquad x + 4 = 0 \Rightarrow x = -4
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=−7±12x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-7 \pm \sqrt{1}}{2}
  7. 7

    Simplify to get the solutions:

    x=−3orx=−4x = -3\quad\text{or}\quad x = -4
  8. 8

    Check by putting x = -3 back in: a·x² + b·x + c = 1·(9) + 7·(-3) + 12 = 0 ✓

LaTeX source of the formulas on this page
\[ x^2 + 7x + 12 = 0\ \Rightarrow\ x = -3,\ -4 \]
\[ x^2 + 7x + 12 = 0 \]
\[ D = 7^2 - 4\cdot1\cdot12 = 49 - 48 = 1 \]
\[ \left(x + 3\right)\left(x + 4\right) = 0 \]
\[ x + 3 = 0 \Rightarrow x = -3\qquad x + 4 = 0 \Rightarrow x = -4 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-7 \pm \sqrt{1}}{2} \]
\[ x = -3\quad\text{or}\quad x = -4 \]

The parabola y = x² + 7x + 12

xy
-6.58.75
-5.53.75
-4.50.75
-3.5-0.25
-2.50.75
-1.53.75
-0.58.75

Questions

What are the solutions of x² + 7x + 12 = 0?

The solutions of x² + 7x + 12 = 0 are x = -3 or x = -4.

Does x² + 7x + 12 = 0 factor?

Yes: (x+3)(x+4) = 0.

What is the vertex of y = x² + 7x + 12?

(-7/2, -1/4). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -7, product = c/a = 12.

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