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Solve 3x² + 4x + 7 = 0

3x² + 4x + 7 = 0 has no real solutions. The two complex solutions are x = (-2 + √17 i)/3 and x = (-2 - √17 i)/3.

x² +x += 0
3x2+4x+7=0⇒x=−2+17 i3, −2−17 i3 (complex)3x^2 + 4x + 7 = 0 \Rightarrow x = \frac{-2 + \sqrt{17}\, i}{3},\ \frac{-2 - \sqrt{17}\, i}{3}\ \text{(complex)}
Discriminant D = -68

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

Graph of y = 3x² + 4x + 7-4-3-2-11231020304050vertexy = 3x² + 4x + 7
Graph of y = 3x² + 4x + 7. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-0.66666667, 5.6666667).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 3, b = 4, c = 7.

    3x2+4x+7=03x^2 + 4x + 7 = 0
  2. 2

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

    D=42−4⋅3⋅7=16−84=−68D = 4^2 - 4\cdot3\cdot7 = 16 - 84 = -68
  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=−4±−686x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{-68}}{6}
  5. 5

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

    x=−2+17 i3orx=−2−17 i3x = \frac{-2 + \sqrt{17}\, i}{3}\quad\text{or}\quad x = \frac{-2 - \sqrt{17}\, i}{3}
LaTeX source of the formulas on this page
\[ 3x^2 + 4x + 7 = 0\ \Rightarrow\ x = \frac{-2 + \sqrt{17}\, i}{3},\ \frac{-2 - \sqrt{17}\, i}{3} \]
\[ 3x^2 + 4x + 7 = 0 \]
\[ D = 4^2 - 4\cdot3\cdot7 = 16 - 84 = -68 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{-68}}{6} \]
\[ x = \frac{-2 + \sqrt{17}\, i}{3}\quad\text{or}\quad x = \frac{-2 - \sqrt{17}\, i}{3} \]

The parabola y = 3x² + 4x + 7

xy
-3.666732.667267
-2.666717.667067
-1.66678.6668667
-0.66675.6666667
0.33338.6664667
1.333317.666267
2.333332.666067

Questions

What are the solutions of 3x² + 4x + 7 = 0?

3x² + 4x + 7 = 0 has no real solutions. The two complex solutions are x = (-2 + √17 i)/3 and x = (-2 - √17 i)/3.

Does 3x² + 4x + 7 = 0 factor?

Not over the integers, because the discriminant -68 is negative. Use the quadratic formula.

What is the vertex of y = 3x² + 4x + 7?

(-2/3, 17/3). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -4/3, product = c/a = 7/3.

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