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Solve 3x² + 5x + 4 = 0

3x² + 5x + 4 = 0 has no real solutions. The two complex solutions are x = (-5 + √23 i)/6 and x = (-5 - √23 i)/6.

x² +x += 0
3x2+5x+4=0⇒x=−5+23 i6, −5−23 i6 (complex)3x^2 + 5x + 4 = 0 \Rightarrow x = \frac{-5 + \sqrt{23}\, i}{6},\ \frac{-5 - \sqrt{23}\, i}{6}\ \text{(complex)}
Discriminant D = -23

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

Graph of y = 3x² + 5x + 4-4-3-2-11231020304050vertexy = 3x² + 5x + 4
Graph of y = 3x² + 5x + 4. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-0.83333333, 1.9166667).

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 = 5, c = 4.

    3x2+5x+4=03x^2 + 5x + 4 = 0
  2. 2

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

    D=52−4⋅3⋅4=25−48=−23D = 5^2 - 4\cdot3\cdot4 = 25 - 48 = -23
  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=−5±−236x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{-23}}{6}
  5. 5

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

    x=−5+23 i6orx=−5−23 i6x = \frac{-5 + \sqrt{23}\, i}{6}\quad\text{or}\quad x = \frac{-5 - \sqrt{23}\, i}{6}
LaTeX source of the formulas on this page
\[ 3x^2 + 5x + 4 = 0\ \Rightarrow\ x = \frac{-5 + \sqrt{23}\, i}{6},\ \frac{-5 - \sqrt{23}\, i}{6} \]
\[ 3x^2 + 5x + 4 = 0 \]
\[ D = 5^2 - 4\cdot3\cdot4 = 25 - 48 = -23 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{-23}}{6} \]
\[ x = \frac{-5 + \sqrt{23}\, i}{6}\quad\text{or}\quad x = \frac{-5 - \sqrt{23}\, i}{6} \]

The parabola y = 3x² + 5x + 4

xy
-3.833328.916067
-2.833313.916267
-1.83334.9164667
-0.83331.9166667
0.16674.9168667
1.166713.917067
2.166728.917267

Questions

What are the solutions of 3x² + 5x + 4 = 0?

3x² + 5x + 4 = 0 has no real solutions. The two complex solutions are x = (-5 + √23 i)/6 and x = (-5 - √23 i)/6.

Does 3x² + 5x + 4 = 0 factor?

Not over the integers, because the discriminant -23 is negative. Use the quadratic formula.

What is the vertex of y = 3x² + 5x + 4?

(-5/6, 23/12). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -5/3, product = c/a = 4/3.

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