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Solve x² - 6x - 5 = 0

The solutions of x² - 6x - 5 = 0 are x = 3 + √14 or x = 3 - √14, that is x ≈ 6.7416574 and x ≈ -0.74165739.

x² +x += 0
x2−6x−5=0⇒x=3+14, 3−14x^2 - 6x - 5 = 0 \Rightarrow x = 3 + \sqrt{14},\ 3 - \sqrt{14}
Discriminant D = 56 · x ≈ 6.74166, -0.741657

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

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Graph of y = x² - 6x - 5-22468-101020x = 6.7416574x = -0.74165739vertexy = x² - 6x - 5
Graph of y = x² - 6x - 5. The curve crosses the x-axis at x = 6.7416574 and x = -0.74165739; the vertex is at (3, -14).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 1, b = -6, c = -5.

    x2−6x−5=0x^2 - 6x - 5 = 0
  2. 2

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

    D=(−6)2−4⋅1⋅(−5)=36+20=56D = (-6)^2 - 4\cdot1\cdot(-5) = 36 + 20 = 56
  3. 3

    D is positive, so there are two different real solutions.

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=6±562x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{56}}{2}
  5. 5

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

    x=3+14orx=3−14x = 3 + \sqrt{14}\quad\text{or}\quad x = 3 - \sqrt{14}
  6. 6

    As decimals: x ≈ 6.7416574 and x ≈ -0.74165739.

  7. 7

    Check numerically: for x ≈ 6.7416574 the left side is -7.1054274e-15 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ x^2 - 6x - 5 = 0\ \Rightarrow\ x = 3 + \sqrt{14},\ 3 - \sqrt{14} \]
\[ x^2 - 6x - 5 = 0 \]
\[ D = (-6)^2 - 4\cdot1\cdot(-5) = 36 + 20 = 56 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{56}}{2} \]
\[ x = 3 + \sqrt{14}\quad\text{or}\quad x = 3 - \sqrt{14} \]

The parabola y = x² - 6x - 5

xy
0-5
1-10
2-13
3-14
4-13
5-10
6-5

Questions

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

The solutions of x² - 6x - 5 = 0 are x = 3 + √14 or x = 3 - √14, that is x ≈ 6.7416574 and x ≈ -0.74165739.

Does x² - 6x - 5 = 0 factor?

Not over the integers, because the discriminant 56 is not a perfect square. Use the quadratic formula.

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

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

What are the sum and product of the roots?

Sum = −b/a = 6, product = c/a = -5.

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