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Solve x² - 12x - 3 = 0

The solutions of x² - 12x - 3 = 0 are x = 6 + √39 or x = 6 - √39, that is x ≈ 12.244998 and x ≈ -0.244998.

x² +x += 0
x2−12x−3=0⇒x=6+39, 6−39x^2 - 12x - 3 = 0 \Rightarrow x = 6 + \sqrt{39},\ 6 - \sqrt{39}
Discriminant D = 156 · x ≈ 12.245, -0.244998

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

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Graph of y = x² - 12x - 3-22468101214-40-202040x = 12.244998x = -0.244998vertexy = x² - 12x - 3
Graph of y = x² - 12x - 3. The curve crosses the x-axis at x = 12.244998 and x = -0.244998; the vertex is at (6, -39).

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 = -12, c = -3.

    x2−12x−3=0x^2 - 12x - 3 = 0
  2. 2

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

    D=(−12)2−4⋅1⋅(−3)=144+12=156D = (-12)^2 - 4\cdot1\cdot(-3) = 144 + 12 = 156
  3. 3

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

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=12±1562x = \frac{-b \pm \sqrt{D}}{2a} = \frac{12 \pm \sqrt{156}}{2}
  5. 5

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

    x=6+39orx=6−39x = 6 + \sqrt{39}\quad\text{or}\quad x = 6 - \sqrt{39}
  6. 6

    As decimals: x ≈ 12.244998 and x ≈ -0.244998.

  7. 7

    Check numerically: for x ≈ 12.244998 the left side is 0 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ x^2 - 12x - 3 = 0\ \Rightarrow\ x = 6 + \sqrt{39},\ 6 - \sqrt{39} \]
\[ x^2 - 12x - 3 = 0 \]
\[ D = (-12)^2 - 4\cdot1\cdot(-3) = 144 + 12 = 156 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{12 \pm \sqrt{156}}{2} \]
\[ x = 6 + \sqrt{39}\quad\text{or}\quad x = 6 - \sqrt{39} \]

The parabola y = x² - 12x - 3

xy
3-30
4-35
5-38
6-39
7-38
8-35
9-30

Questions

What are the solutions of x² - 12x - 3 = 0?

The solutions of x² - 12x - 3 = 0 are x = 6 + √39 or x = 6 - √39, that is x ≈ 12.244998 and x ≈ -0.244998.

Does x² - 12x - 3 = 0 factor?

Not over the integers, because the discriminant 156 is not a perfect square. Use the quadratic formula.

What is the vertex of y = x² - 12x - 3?

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

What are the sum and product of the roots?

Sum = −b/a = 12, product = c/a = -3.

Similar equations

    Solve x² - 12x - 3 = 0: x = 6 + √39 or x = 6 - √39 (steps) | WordGPT