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Solve 2x² + 6x = 0

The solutions of 2x² + 6x = 0 are x = 0 or x = -3.

x² +x += 0
2x2+6x=0⇒x=0, −32x^2 + 6x = 0 \Rightarrow x = 0,\ -3
Discriminant D = 36 · x ≈ 0, -3

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

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Graph of y = 2x² + 6x-5-4-3-2-112-551015202530x = 0x = -3vertexy = 2x² + 6x
Graph of y = 2x² + 6x. The curve crosses the x-axis at x = 0 and x = -3; the vertex is at (-1.5, -4.5).

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 = 6, c = 0.

    2x2+6x=02x^2 + 6x = 0
  2. 2

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

    D=62−4⋅2⋅0=36+0=36D = 6^2 - 4\cdot2\cdot0 = 36 + 0 = 36
  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+3)=02\left(x\right)\left(x + 3\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+3=0⇒x=−3x = 0 \Rightarrow x = 0\qquad x + 3 = 0 \Rightarrow x = -3
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=−6±364x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{36}}{4}
  7. 7

    Simplify to get the solutions:

    x=0orx=−3x = 0\quad\text{or}\quad x = -3
  8. 8

    Check by putting x = 0 back in: a·x² + b·x + c = 2·(0) + 6·(0) + 0 = 0 ✓

LaTeX source of the formulas on this page
\[ 2x^2 + 6x = 0\ \Rightarrow\ x = 0,\ -3 \]
\[ 2x^2 + 6x = 0 \]
\[ D = 6^2 - 4\cdot2\cdot0 = 36 + 0 = 36 \]
\[ 2\left(x\right)\left(x + 3\right) = 0 \]
\[ x = 0 \Rightarrow x = 0\qquad x + 3 = 0 \Rightarrow x = -3 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{36}}{4} \]
\[ x = 0\quad\text{or}\quad x = -3 \]

The parabola y = 2x² + 6x

xy
-4.513.5
-3.53.5
-2.5-2.5
-1.5-4.5
-0.5-2.5
0.53.5
1.513.5

Questions

What are the solutions of 2x² + 6x = 0?

The solutions of 2x² + 6x = 0 are x = 0 or x = -3.

Does 2x² + 6x = 0 factor?

Yes: 2(x)(x+3) = 0.

What is the vertex of y = 2x² + 6x?

(-3/2, -9/2). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -3, product = c/a = 0.

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