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Solve 2x² + 12x + 10 = 0

The solutions of 2x² + 12x + 10 = 0 are x = -1 or x = -5.

x² +x += 0
2x2+12x+10=0⇒x=−1, −52x^2 + 12x + 10 = 0 \Rightarrow x = -1,\ -5
Discriminant D = 64 · x ≈ -1, -5

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

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Graph of y = 2x² + 12x + 10-7-6-5-4-3-2-11-10102030x = -1x = -5vertexy = 2x² + 12x + 10
Graph of y = 2x² + 12x + 10. The curve crosses the x-axis at x = -1 and x = -5; the vertex is at (-3, -8).

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 = 10.

    2x2+12x+10=02x^2 + 12x + 10 = 0
  2. 2

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

    D=122−4⋅2⋅10=144−80=64D = 12^2 - 4\cdot2\cdot10 = 144 - 80 = 64
  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+1)(x+5)=02\left(x + 1\right)\left(x + 5\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+1=0⇒x=−1x+5=0⇒x=−5x + 1 = 0 \Rightarrow x = -1\qquad x + 5 = 0 \Rightarrow x = -5
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=−12±644x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{64}}{4}
  7. 7

    Simplify to get the solutions:

    x=−1orx=−5x = -1\quad\text{or}\quad x = -5
  8. 8

    Check by putting x = -1 back in: a·x² + b·x + c = 2·(1) + 12·(-1) + 10 = 0 ✓

LaTeX source of the formulas on this page
\[ 2x^2 + 12x + 10 = 0\ \Rightarrow\ x = -1,\ -5 \]
\[ 2x^2 + 12x + 10 = 0 \]
\[ D = 12^2 - 4\cdot2\cdot10 = 144 - 80 = 64 \]
\[ 2\left(x + 1\right)\left(x + 5\right) = 0 \]
\[ x + 1 = 0 \Rightarrow x = -1\qquad x + 5 = 0 \Rightarrow x = -5 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{64}}{4} \]
\[ x = -1\quad\text{or}\quad x = -5 \]

The parabola y = 2x² + 12x + 10

xy
-610
-50
-4-6
-3-8
-2-6
-10
010

Questions

What are the solutions of 2x² + 12x + 10 = 0?

The solutions of 2x² + 12x + 10 = 0 are x = -1 or x = -5.

Does 2x² + 12x + 10 = 0 factor?

Yes: 2(x+1)(x+5) = 0.

What is the vertex of y = 2x² + 12x + 10?

(-3, -8). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -6, product = c/a = 5.

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