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Solve -2x² + 12x = 0

The solutions of -2x² + 12x = 0 are x = 0 or x = 6.

x² +x += 0
−2x2+12x=0⇒x=0, 6-2x^2 + 12x = 0 \Rightarrow x = 0,\ 6
Discriminant D = 144 · x ≈ 0, 6

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

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Graph of y = -2x² + 12x-22468-40-30-20-101020x = 0x = 6vertexy = -2x² + 12x
Graph of y = -2x² + 12x. The curve crosses the x-axis at x = 0 and x = 6; the vertex is at (3, 18).

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 = 0.

    −2x2+12x=0-2x^2 + 12x = 0
  2. 2

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

    D=122−4⋅(−2)⋅0=144+0=144D = 12^2 - 4\cdot(-2)\cdot0 = 144 + 0 = 144
  3. 3

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

  4. 4

    Method 1, factoring. The roots are rational, so the left side factors:

    −2(x)(x−6)=0-2\left(x\right)\left(x - 6\right) = 0
  5. 5

    A product is zero when one of its factors is zero, so set each factor equal to 0 and solve:

    x=0⇒x=0x−6=0⇒x=6x = 0 \Rightarrow x = 0\qquad x - 6 = 0 \Rightarrow x = 6
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=−12±144−4x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{144}}{-4}
  7. 7

    Simplify to get the solutions:

    x=0orx=6x = 0\quad\text{or}\quad x = 6
  8. 8

    Check by putting x = 0 back in: a·x² + b·x + c = -2·(0) + 12·(0) + 0 = 0 ✓

LaTeX source of the formulas on this page
\[ -2x^2 + 12x = 0\ \Rightarrow\ x = 0,\ 6 \]
\[ -2x^2 + 12x = 0 \]
\[ D = 12^2 - 4\cdot(-2)\cdot0 = 144 + 0 = 144 \]
\[ -2\left(x\right)\left(x - 6\right) = 0 \]
\[ x = 0 \Rightarrow x = 0\qquad x - 6 = 0 \Rightarrow x = 6 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{144}}{-4} \]
\[ x = 0\quad\text{or}\quad x = 6 \]

The parabola y = -2x² + 12x

xy
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Questions

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

The solutions of -2x² + 12x = 0 are x = 0 or x = 6.

Does -2x² + 12x = 0 factor?

Yes: -2(x)(x-6) = 0.

What is the vertex of y = -2x² + 12x?

(3, 18). 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 = 0.

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