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Solve 2x² + 5x + 7 = 0

2x² + 5x + 7 = 0 has no real solutions. The two complex solutions are x = (-5 + √31 i)/4 and x = (-5 - √31 i)/4.

x² +x += 0
2x2+5x+7=0⇒x=−5+31 i4, −5−31 i4 (complex)2x^2 + 5x + 7 = 0 \Rightarrow x = \frac{-5 + \sqrt{31}\, i}{4},\ \frac{-5 - \sqrt{31}\, i}{4}\ \text{(complex)}
Discriminant D = -31

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

Graph of y = 2x² + 5x + 7-5-4-3-2-1125101520253035vertexy = 2x² + 5x + 7
Graph of y = 2x² + 5x + 7. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-1.25, 3.875).

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 = 5, c = 7.

    2x2+5x+7=02x^2 + 5x + 7 = 0
  2. 2

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

    D=52−4⋅2⋅7=25−56=−31D = 5^2 - 4\cdot2\cdot7 = 25 - 56 = -31
  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±−314x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{-31}}{4}
  5. 5

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

    x=−5+31 i4orx=−5−31 i4x = \frac{-5 + \sqrt{31}\, i}{4}\quad\text{or}\quad x = \frac{-5 - \sqrt{31}\, i}{4}
LaTeX source of the formulas on this page
\[ 2x^2 + 5x + 7 = 0\ \Rightarrow\ x = \frac{-5 + \sqrt{31}\, i}{4},\ \frac{-5 - \sqrt{31}\, i}{4} \]
\[ 2x^2 + 5x + 7 = 0 \]
\[ D = 5^2 - 4\cdot2\cdot7 = 25 - 56 = -31 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{-31}}{4} \]
\[ x = \frac{-5 + \sqrt{31}\, i}{4}\quad\text{or}\quad x = \frac{-5 - \sqrt{31}\, i}{4} \]

The parabola y = 2x² + 5x + 7

xy
-4.2521.875
-3.2511.875
-2.255.875
-1.253.875
-0.255.875
0.7511.875
1.7521.875

Questions

What are the solutions of 2x² + 5x + 7 = 0?

2x² + 5x + 7 = 0 has no real solutions. The two complex solutions are x = (-5 + √31 i)/4 and x = (-5 - √31 i)/4.

Does 2x² + 5x + 7 = 0 factor?

Not over the integers, because the discriminant -31 is negative. Use the quadratic formula.

What is the vertex of y = 2x² + 5x + 7?

(-5/4, 31/8). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -5/2, product = c/a = 7/2.

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