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Solve x² - 3x + 12 = 0

x² - 3x + 12 = 0 has no real solutions. The two complex solutions are x = (3 + √39 i)/2 and x = (3 - √39 i)/2.

x² +x += 0
x2−3x+12=0⇒x=3+39 i2, 3−39 i2 (complex)x^2 - 3x + 12 = 0 \Rightarrow x = \frac{3 + \sqrt{39}\, i}{2},\ \frac{3 - \sqrt{39}\, i}{2}\ \text{(complex)}
Discriminant D = -39

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

Graph of y = x² - 3x + 12-2-1123458101214161820222426vertexy = x² - 3x + 12
Graph of y = x² - 3x + 12. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (1.5, 9.75).

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 = -3, c = 12.

    x2−3x+12=0x^2 - 3x + 12 = 0
  2. 2

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

    D=(−3)2−4⋅1⋅12=9−48=−39D = (-3)^2 - 4\cdot1\cdot12 = 9 - 48 = -39
  3. 3

    D is negative, so there are no real solutions; the two solutions are complex numbers.

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=3±−392x = \frac{-b \pm \sqrt{D}}{2a} = \frac{3 \pm \sqrt{-39}}{2}
  5. 5

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

    x=3+39 i2orx=3−39 i2x = \frac{3 + \sqrt{39}\, i}{2}\quad\text{or}\quad x = \frac{3 - \sqrt{39}\, i}{2}
LaTeX source of the formulas on this page
\[ x^2 - 3x + 12 = 0\ \Rightarrow\ x = \frac{3 + \sqrt{39}\, i}{2},\ \frac{3 - \sqrt{39}\, i}{2} \]
\[ x^2 - 3x + 12 = 0 \]
\[ D = (-3)^2 - 4\cdot1\cdot12 = 9 - 48 = -39 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{3 \pm \sqrt{-39}}{2} \]
\[ x = \frac{3 + \sqrt{39}\, i}{2}\quad\text{or}\quad x = \frac{3 - \sqrt{39}\, i}{2} \]

The parabola y = x² - 3x + 12

xy
-1.518.75
-0.513.75
0.510.75
1.59.75
2.510.75
3.513.75
4.518.75

Questions

What are the solutions of x² - 3x + 12 = 0?

x² - 3x + 12 = 0 has no real solutions. The two complex solutions are x = (3 + √39 i)/2 and x = (3 - √39 i)/2.

Does x² - 3x + 12 = 0 factor?

Not over the integers, because the discriminant -39 is negative. Use the quadratic formula.

What is the vertex of y = x² - 3x + 12?

(3/2, 39/4). 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 = 12.

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