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Solve 5x² + 12x + 10 = 0

5x² + 12x + 10 = 0 has no real solutions. The two complex solutions are x = (-6 + √14 i)/5 and x = (-6 - √14 i)/5.

x² +x += 0
5x2+12x+10=0⇒x=−6+14 i5, −6−14 i5 (complex)5x^2 + 12x + 10 = 0 \Rightarrow x = \frac{-6 + \sqrt{14}\, i}{5},\ \frac{-6 - \sqrt{14}\, i}{5}\ \text{(complex)}
Discriminant D = -56

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

Graph of y = 5x² + 12x + 10-5-4-3-2-11220406080vertexy = 5x² + 12x + 10
Graph of y = 5x² + 12x + 10. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-1.2, 2.8).

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 = 12, c = 10.

    5x2+12x+10=05x^2 + 12x + 10 = 0
  2. 2

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

    D=122−4⋅5⋅10=144−200=−56D = 12^2 - 4\cdot5\cdot10 = 144 - 200 = -56
  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=−12±−5610x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{-56}}{10}
  5. 5

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

    x=−6+14 i5orx=−6−14 i5x = \frac{-6 + \sqrt{14}\, i}{5}\quad\text{or}\quad x = \frac{-6 - \sqrt{14}\, i}{5}
LaTeX source of the formulas on this page
\[ 5x^2 + 12x + 10 = 0\ \Rightarrow\ x = \frac{-6 + \sqrt{14}\, i}{5},\ \frac{-6 - \sqrt{14}\, i}{5} \]
\[ 5x^2 + 12x + 10 = 0 \]
\[ D = 12^2 - 4\cdot5\cdot10 = 144 - 200 = -56 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{-56}}{10} \]
\[ x = \frac{-6 + \sqrt{14}\, i}{5}\quad\text{or}\quad x = \frac{-6 - \sqrt{14}\, i}{5} \]

The parabola y = 5x² + 12x + 10

xy
-4.247.8
-3.222.8
-2.27.8
-1.22.8
-0.27.8
0.822.8
1.847.8

Questions

What are the solutions of 5x² + 12x + 10 = 0?

5x² + 12x + 10 = 0 has no real solutions. The two complex solutions are x = (-6 + √14 i)/5 and x = (-6 - √14 i)/5.

Does 5x² + 12x + 10 = 0 factor?

Not over the integers, because the discriminant -56 is negative. Use the quadratic formula.

What is the vertex of y = 5x² + 12x + 10?

(-6/5, 14/5). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

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

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