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Solve x² + 12x - 5 = 0

The solutions of x² + 12x - 5 = 0 are x = -6 + √41 or x = -6 - √41, that is x ≈ 0.40312424 and x ≈ -12.403124.

x² +x += 0
x2+12x−5=0⇒x=−6+41, −6−41x^2 + 12x - 5 = 0 \Rightarrow x = -6 + \sqrt{41},\ -6 - \sqrt{41}
Discriminant D = 164 · x ≈ 0.403124, -12.4031

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

Graph of y = x² + 12x - 5-14-12-10-8-6-4-22-40-202040x = 0.40312424x = -12.403124vertexy = x² + 12x - 5
Graph of y = x² + 12x - 5. The curve crosses the x-axis at x = 0.40312424 and x = -12.403124; the vertex is at (-6, -41).

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 = -5.

    x2+12x−5=0x^2 + 12x - 5 = 0
  2. 2

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

    D=122−4⋅1⋅(−5)=144+20=164D = 12^2 - 4\cdot1\cdot(-5) = 144 + 20 = 164
  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±1642x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{164}}{2}
  5. 5

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

    x=−6+41orx=−6−41x = -6 + \sqrt{41}\quad\text{or}\quad x = -6 - \sqrt{41}
  6. 6

    As decimals: x ≈ 0.40312424 and x ≈ -12.403124.

  7. 7

    Check numerically: for x ≈ 0.40312424 the left side is -1.7763568e-15 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ x^2 + 12x - 5 = 0\ \Rightarrow\ x = -6 + \sqrt{41},\ -6 - \sqrt{41} \]
\[ x^2 + 12x - 5 = 0 \]
\[ D = 12^2 - 4\cdot1\cdot(-5) = 144 + 20 = 164 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{164}}{2} \]
\[ x = -6 + \sqrt{41}\quad\text{or}\quad x = -6 - \sqrt{41} \]

The parabola y = x² + 12x - 5

xy
-9-32
-8-37
-7-40
-6-41
-5-40
-4-37
-3-32

Questions

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

The solutions of x² + 12x - 5 = 0 are x = -6 + √41 or x = -6 - √41, that is x ≈ 0.40312424 and x ≈ -12.403124.

Does x² + 12x - 5 = 0 factor?

Not over the integers, because the discriminant 164 is not a perfect square. Use the quadratic formula.

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

(-6, -41). 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 = -5.

Similar equations

    Solve x² + 12x - 5 = 0: x = -6 + √41 or x = -6 - √41 (steps) | WordGPT