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Solve -2x² + 12x - 5 = 0

The solutions of -2x² + 12x - 5 = 0 are x = (6 + √26)/2 or x = (6 - √26)/2, that is x ≈ 0.45049024 and x ≈ 5.5495098.

x² +x += 0
−2x2+12x−5=0⇒x=6+262, 6−262-2x^2 + 12x - 5 = 0 \Rightarrow x = \frac{6 + \sqrt{26}}{2},\ \frac{6 - \sqrt{26}}{2}
Discriminant D = 104 · x ≈ 0.45049, 5.54951

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

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Graph of y = -2x² + 12x - 5-2-112345678-40-30-20-1010x = 0.45049024x = 5.5495098vertexy = -2x² + 12x - 5
Graph of y = -2x² + 12x - 5. The curve crosses the x-axis at x = 0.45049024 and x = 5.5495098; 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 = -2, b = 12, c = -5.

    −2x2+12x−5=0-2x^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⋅(−2)⋅(−5)=144−40=104D = 12^2 - 4\cdot(-2)\cdot(-5) = 144 - 40 = 104
  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±104−4x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{104}}{-4}
  5. 5

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

    x=6+262orx=6−262x = \frac{6 + \sqrt{26}}{2}\quad\text{or}\quad x = \frac{6 - \sqrt{26}}{2}
  6. 6

    As decimals: x ≈ 0.45049024 and x ≈ 5.5495098.

  7. 7

    Check numerically: for x ≈ 0.45049024 the left side is 1.7763568e-15 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ -2x^2 + 12x - 5 = 0\ \Rightarrow\ x = \frac{6 + \sqrt{26}}{2},\ \frac{6 - \sqrt{26}}{2} \]
\[ -2x^2 + 12x - 5 = 0 \]
\[ D = 12^2 - 4\cdot(-2)\cdot(-5) = 144 - 40 = 104 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{104}}{-4} \]
\[ x = \frac{6 + \sqrt{26}}{2}\quad\text{or}\quad x = \frac{6 - \sqrt{26}}{2} \]

The parabola y = -2x² + 12x - 5

xy
0-5
15
211
313
411
55
6-5

Questions

What are the solutions of -2x² + 12x - 5 = 0?

The solutions of -2x² + 12x - 5 = 0 are x = (6 + √26)/2 or x = (6 - √26)/2, that is x ≈ 0.45049024 and x ≈ 5.5495098.

Does -2x² + 12x - 5 = 0 factor?

Not over the integers, because the discriminant 104 is not a perfect square. Use the quadratic formula.

What is the vertex of y = -2x² + 12x - 5?

(3, 13). 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 = 5/2.

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    Solve -2x² + 12x - 5 = 0: x = (6 + √26)/2 or x = (6 - √26)/2 (steps) | WordGPT