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Solve -x² + 6x + 1 = 0

The solutions of -x² + 6x + 1 = 0 are x = 3 + √10 or x = 3 - √10, that is x ≈ -0.16227766 and x ≈ 6.1622777.

x² +x += 0
−x2+6x+1=0⇒x=3+10, 3−10-x^2 + 6x + 1 = 0 \Rightarrow x = 3 + \sqrt{10},\ 3 - \sqrt{10}
Discriminant D = 40 · x ≈ -0.162278, 6.16228

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

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Graph of y = -x² + 6x + 1-22468-20-15-10-5510x = -0.16227766x = 6.1622777vertexy = -x² + 6x + 1
Graph of y = -x² + 6x + 1. The curve crosses the x-axis at x = -0.16227766 and x = 6.1622777; the vertex is at (3, 10).

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 = 1.

    −x2+6x+1=0-x^2 + 6x + 1 = 0
  2. 2

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

    D=62−4⋅(−1)⋅1=36+4=40D = 6^2 - 4\cdot(-1)\cdot1 = 36 + 4 = 40
  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±40−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{40}}{-2}
  5. 5

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

    x=3+10orx=3−10x = 3 + \sqrt{10}\quad\text{or}\quad x = 3 - \sqrt{10}
  6. 6

    As decimals: x ≈ -0.16227766 and x ≈ 6.1622777.

  7. 7

    Check numerically: for x ≈ -0.16227766 the left side is -1.110223e-15 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ -x^2 + 6x + 1 = 0\ \Rightarrow\ x = 3 + \sqrt{10},\ 3 - \sqrt{10} \]
\[ -x^2 + 6x + 1 = 0 \]
\[ D = 6^2 - 4\cdot(-1)\cdot1 = 36 + 4 = 40 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{40}}{-2} \]
\[ x = 3 + \sqrt{10}\quad\text{or}\quad x = 3 - \sqrt{10} \]

The parabola y = -x² + 6x + 1

xy
01
16
29
310
49
56
61

Questions

What are the solutions of -x² + 6x + 1 = 0?

The solutions of -x² + 6x + 1 = 0 are x = 3 + √10 or x = 3 - √10, that is x ≈ -0.16227766 and x ≈ 6.1622777.

Does -x² + 6x + 1 = 0 factor?

Not over the integers, because the discriminant 40 is not a perfect square. Use the quadratic formula.

What is the vertex of y = -x² + 6x + 1?

(3, 10). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 6, product = c/a = -1.

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