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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 ≈ -0.244998 and x ≈ 12.244998.

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

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 = -0.244998x = 12.244998vertexy = -x² + 12x + 3
Graph of y = -x² + 12x + 3. The curve crosses the x-axis at x = -0.244998 and x = 12.244998; 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=0-x^2 + 12x + 3 = 0
  2. 2

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

    D=122−4⋅(−1)⋅3=144+12=156D = 12^2 - 4\cdot(-1)\cdot3 = 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±156−2x = \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 ≈ -0.244998 and x ≈ 12.244998.

  7. 7

    Check numerically: for x ≈ -0.244998 the left side is -1.3322676e-15 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ -x^2 + 12x + 3 = 0\ \Rightarrow\ x = 6 + \sqrt{39},\ 6 - \sqrt{39} \]
\[ -x^2 + 12x + 3 = 0 \]
\[ D = 12^2 - 4\cdot(-1)\cdot3 = 144 + 12 = 156 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{156}}{-2} \]
\[ x = 6 + \sqrt{39}\quad\text{or}\quad x = 6 - \sqrt{39} \]

The parabola y = -x² + 12x + 3

xy
330
435
538
639
738
835
930

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 ≈ -0.244998 and x ≈ 12.244998.

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 downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 12, product = c/a = -3.

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