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Solve 5x² + 10x + 6 = 0

5x² + 10x + 6 = 0 has no real solutions. The two complex solutions are x = (-5 + √5 i)/5 and x = (-5 - √5 i)/5.

x² +x += 0
5x2+10x+6=0⇒x=−5+5 i5, −5−5 i5 (complex)5x^2 + 10x + 6 = 0 \Rightarrow x = \frac{-5 + \sqrt{5}\, i}{5},\ \frac{-5 - \sqrt{5}\, i}{5}\ \text{(complex)}
Discriminant D = -20

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

Graph of y = 5x² + 10x + 6-5-4-3-2-112320406080vertexy = 5x² + 10x + 6
Graph of y = 5x² + 10x + 6. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-1, 1).

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 = 10, c = 6.

    5x2+10x+6=05x^2 + 10x + 6 = 0
  2. 2

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

    D=102−4⋅5⋅6=100−120=−20D = 10^2 - 4\cdot5\cdot6 = 100 - 120 = -20
  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=−10±−2010x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-10 \pm \sqrt{-20}}{10}
  5. 5

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

    x=−5+5 i5orx=−5−5 i5x = \frac{-5 + \sqrt{5}\, i}{5}\quad\text{or}\quad x = \frac{-5 - \sqrt{5}\, i}{5}
LaTeX source of the formulas on this page
\[ 5x^2 + 10x + 6 = 0\ \Rightarrow\ x = \frac{-5 + \sqrt{5}\, i}{5},\ \frac{-5 - \sqrt{5}\, i}{5} \]
\[ 5x^2 + 10x + 6 = 0 \]
\[ D = 10^2 - 4\cdot5\cdot6 = 100 - 120 = -20 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-10 \pm \sqrt{-20}}{10} \]
\[ x = \frac{-5 + \sqrt{5}\, i}{5}\quad\text{or}\quad x = \frac{-5 - \sqrt{5}\, i}{5} \]

The parabola y = 5x² + 10x + 6

xy
-446
-321
-26
-11
06
121
246

Questions

What are the solutions of 5x² + 10x + 6 = 0?

5x² + 10x + 6 = 0 has no real solutions. The two complex solutions are x = (-5 + √5 i)/5 and x = (-5 - √5 i)/5.

Does 5x² + 10x + 6 = 0 factor?

Not over the integers, because the discriminant -20 is negative. Use the quadratic formula.

What is the vertex of y = 5x² + 10x + 6?

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

What are the sum and product of the roots?

Sum = −b/a = -2, product = c/a = 6/5.

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