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Solve x² - 6x - 4 = 0

The solutions of x² - 6x - 4 = 0 are x = 3 + √13 or x = 3 - √13, that is x ≈ 6.6055513 and x ≈ -0.60555128.

x² +x += 0
x2−6x−4=0⇒x=3+13, 3−13x^2 - 6x - 4 = 0 \Rightarrow x = 3 + \sqrt{13},\ 3 - \sqrt{13}
Discriminant D = 52 · x ≈ 6.60555, -0.605551

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

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Graph of y = x² - 6x - 4-22468-101020x = 6.6055513x = -0.60555128vertexy = x² - 6x - 4
Graph of y = x² - 6x - 4. The curve crosses the x-axis at x = 6.6055513 and x = -0.60555128; the vertex is at (3, -13).

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 = -4.

    x2−6x−4=0x^2 - 6x - 4 = 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⋅(−4)=36+16=52D = (-6)^2 - 4\cdot1\cdot(-4) = 36 + 16 = 52
  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±522x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{52}}{2}
  5. 5

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

    x=3+13orx=3−13x = 3 + \sqrt{13}\quad\text{or}\quad x = 3 - \sqrt{13}
  6. 6

    As decimals: x ≈ 6.6055513 and x ≈ -0.60555128.

  7. 7

    Check numerically: for x ≈ 6.6055513 the left side is 0 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ x^2 - 6x - 4 = 0\ \Rightarrow\ x = 3 + \sqrt{13},\ 3 - \sqrt{13} \]
\[ x^2 - 6x - 4 = 0 \]
\[ D = (-6)^2 - 4\cdot1\cdot(-4) = 36 + 16 = 52 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{52}}{2} \]
\[ x = 3 + \sqrt{13}\quad\text{or}\quad x = 3 - \sqrt{13} \]

The parabola y = x² - 6x - 4

xy
0-4
1-9
2-12
3-13
4-12
5-9
6-4

Questions

What are the solutions of x² - 6x - 4 = 0?

The solutions of x² - 6x - 4 = 0 are x = 3 + √13 or x = 3 - √13, that is x ≈ 6.6055513 and x ≈ -0.60555128.

Does x² - 6x - 4 = 0 factor?

Not over the integers, because the discriminant 52 is not a perfect square. Use the quadratic formula.

What is the vertex of y = x² - 6x - 4?

(3, -13). 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 = -4.

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