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Solve 2x² - x + 5 = 0

2x² - x + 5 = 0 has no real solutions. The two complex solutions are x = (1 + √39 i)/4 and x = (1 - √39 i)/4.

x² +x += 0
2x2−x+5=0⇒x=1+39 i4, 1−39 i4 (complex)2x^2 - x + 5 = 0 \Rightarrow x = \frac{1 + \sqrt{39}\, i}{4},\ \frac{1 - \sqrt{39}\, i}{4}\ \text{(complex)}
Discriminant D = -39

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

Graph of y = 2x² - x + 5-3-2-112345101520253035vertexy = 2x² - x + 5
Graph of y = 2x² - x + 5. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0.25, 4.875).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 2, b = -1, c = 5.

    2x2−x+5=02x^2 - x + 5 = 0
  2. 2

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

    D=(−1)2−4⋅2⋅5=1−40=−39D = (-1)^2 - 4\cdot2\cdot5 = 1 - 40 = -39
  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=1±−394x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm \sqrt{-39}}{4}
  5. 5

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

    x=1+39 i4orx=1−39 i4x = \frac{1 + \sqrt{39}\, i}{4}\quad\text{or}\quad x = \frac{1 - \sqrt{39}\, i}{4}
LaTeX source of the formulas on this page
\[ 2x^2 - x + 5 = 0\ \Rightarrow\ x = \frac{1 + \sqrt{39}\, i}{4},\ \frac{1 - \sqrt{39}\, i}{4} \]
\[ 2x^2 - x + 5 = 0 \]
\[ D = (-1)^2 - 4\cdot2\cdot5 = 1 - 40 = -39 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm \sqrt{-39}}{4} \]
\[ x = \frac{1 + \sqrt{39}\, i}{4}\quad\text{or}\quad x = \frac{1 - \sqrt{39}\, i}{4} \]

The parabola y = 2x² - x + 5

xy
-2.7522.875
-1.7512.875
-0.756.875
0.254.875
1.256.875
2.2512.875
3.2522.875

Questions

What are the solutions of 2x² - x + 5 = 0?

2x² - x + 5 = 0 has no real solutions. The two complex solutions are x = (1 + √39 i)/4 and x = (1 - √39 i)/4.

Does 2x² - x + 5 = 0 factor?

Not over the integers, because the discriminant -39 is negative. Use the quadratic formula.

What is the vertex of y = 2x² - x + 5?

(1/4, 39/8). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = 1/2, product = c/a = 5/2.

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