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Solve 5x² + 2x + 12 = 0

5x² + 2x + 12 = 0 has no real solutions. The two complex solutions are x = (-1 + √59 i)/5 and x = (-1 - √59 i)/5.

x² +x += 0
5x2+2x+12=0⇒x=−1+59 i5, −1−59 i5 (complex)5x^2 + 2x + 12 = 0 \Rightarrow x = \frac{-1 + \sqrt{59}\, i}{5},\ \frac{-1 - \sqrt{59}\, i}{5}\ \text{(complex)}
Discriminant D = -236

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

Graph of y = 5x² + 2x + 12-4-3-2-112320406080vertexy = 5x² + 2x + 12
Graph of y = 5x² + 2x + 12. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-0.2, 11.8).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 5, b = 2, c = 12.

    5x2+2x+12=05x^2 + 2x + 12 = 0
  2. 2

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

    D=22−4⋅5⋅12=4−240=−236D = 2^2 - 4\cdot5\cdot12 = 4 - 240 = -236
  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=−2±−23610x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-236}}{10}
  5. 5

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

    x=−1+59 i5orx=−1−59 i5x = \frac{-1 + \sqrt{59}\, i}{5}\quad\text{or}\quad x = \frac{-1 - \sqrt{59}\, i}{5}
LaTeX source of the formulas on this page
\[ 5x^2 + 2x + 12 = 0\ \Rightarrow\ x = \frac{-1 + \sqrt{59}\, i}{5},\ \frac{-1 - \sqrt{59}\, i}{5} \]
\[ 5x^2 + 2x + 12 = 0 \]
\[ D = 2^2 - 4\cdot5\cdot12 = 4 - 240 = -236 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-236}}{10} \]
\[ x = \frac{-1 + \sqrt{59}\, i}{5}\quad\text{or}\quad x = \frac{-1 - \sqrt{59}\, i}{5} \]

The parabola y = 5x² + 2x + 12

xy
-3.256.8
-2.231.8
-1.216.8
-0.211.8
0.816.8
1.831.8
2.856.8

Questions

What are the solutions of 5x² + 2x + 12 = 0?

5x² + 2x + 12 = 0 has no real solutions. The two complex solutions are x = (-1 + √59 i)/5 and x = (-1 - √59 i)/5.

Does 5x² + 2x + 12 = 0 factor?

Not over the integers, because the discriminant -236 is negative. Use the quadratic formula.

What is the vertex of y = 5x² + 2x + 12?

(-1/5, 59/5). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -2/5, product = c/a = 12/5.

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