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Solve x² + 3x + 6 = 0

x² + 3x + 6 = 0 has no real solutions. The two complex solutions are x = (-3 + √15 i)/2 and x = (-3 - √15 i)/2.

x² +x += 0
x2+3x+6=0⇒x=−3+15 i2, −3−15 i2 (complex)x^2 + 3x + 6 = 0 \Rightarrow x = \frac{-3 + \sqrt{15}\, i}{2},\ \frac{-3 - \sqrt{15}\, i}{2}\ \text{(complex)}
Discriminant D = -15

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

Graph of y = x² + 3x + 6-5-4-3-2-1122468101214161820vertexy = x² + 3x + 6
Graph of y = x² + 3x + 6. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-1.5, 3.75).

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 = 3, c = 6.

    x2+3x+6=0x^2 + 3x + 6 = 0
  2. 2

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

    D=32−4⋅1⋅6=9−24=−15D = 3^2 - 4\cdot1\cdot6 = 9 - 24 = -15
  3. 3

    D is negative, so there are no real solutions; the two solutions are complex numbers.

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=−3±−152x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-15}}{2}
  5. 5

    Simplify (pull the square out of the root) to get the solutions:

    x=−3+15 i2orx=−3−15 i2x = \frac{-3 + \sqrt{15}\, i}{2}\quad\text{or}\quad x = \frac{-3 - \sqrt{15}\, i}{2}
LaTeX source of the formulas on this page
\[ x^2 + 3x + 6 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{15}\, i}{2},\ \frac{-3 - \sqrt{15}\, i}{2} \]
\[ x^2 + 3x + 6 = 0 \]
\[ D = 3^2 - 4\cdot1\cdot6 = 9 - 24 = -15 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-15}}{2} \]
\[ x = \frac{-3 + \sqrt{15}\, i}{2}\quad\text{or}\quad x = \frac{-3 - \sqrt{15}\, i}{2} \]

The parabola y = x² + 3x + 6

xy
-4.512.75
-3.57.75
-2.54.75
-1.53.75
-0.54.75
0.57.75
1.512.75

Questions

What are the solutions of x² + 3x + 6 = 0?

x² + 3x + 6 = 0 has no real solutions. The two complex solutions are x = (-3 + √15 i)/2 and x = (-3 - √15 i)/2.

Does x² + 3x + 6 = 0 factor?

Not over the integers, because the discriminant -15 is negative. Use the quadratic formula.

What is the vertex of y = x² + 3x + 6?

(-3/2, 15/4). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -3, product = c/a = 6.

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