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Solve -3x² + 5x + 12 = 0

The solutions of -3x² + 5x + 12 = 0 are x = -4/3 or x = 3.

x² +x += 0
−3x2+5x+12=0⇒x=−43, 3-3x^2 + 5x + 12 = 0 \Rightarrow x = -\frac{4}{3},\ 3
Discriminant D = 169 · x ≈ -1.33333, 3

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

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Graph of y = -3x² + 5x + 12-3-2-112345-50-40-30-20-101020x = -1.3333333x = 3vertexy = -3x² + 5x + 12
Graph of y = -3x² + 5x + 12. The curve crosses the x-axis at x = -1.3333333 and x = 3; the vertex is at (0.83333333, 14.083333).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = -3, b = 5, c = 12.

    −3x2+5x+12=0-3x^2 + 5x + 12 = 0
  2. 2

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

    D=52−4⋅(−3)⋅12=25+144=169D = 5^2 - 4\cdot(-3)\cdot12 = 25 + 144 = 169
  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:

    −(3x+4)(x−3)=0-\left(3x + 4\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:

    3x+4=0⇒x=−43x−3=0⇒x=33x + 4 = 0 \Rightarrow x = -\frac{4}{3}\qquad x - 3 = 0 \Rightarrow x = 3
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=−5±169−6x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{169}}{-6}
  7. 7

    Simplify to get the solutions:

    x=−43orx=3x = -\frac{4}{3}\quad\text{or}\quad x = 3
  8. 8

    Check by putting x = -4/3 back in: a·x² + b·x + c = -3·(16/9) + 5·(-4/3) + 12 = 0 ✓

LaTeX source of the formulas on this page
\[ -3x^2 + 5x + 12 = 0\ \Rightarrow\ x = -\frac{4}{3},\ 3 \]
\[ -3x^2 + 5x + 12 = 0 \]
\[ D = 5^2 - 4\cdot(-3)\cdot12 = 25 + 144 = 169 \]
\[ -\left(3x + 4\right)\left(x - 3\right) = 0 \]
\[ 3x + 4 = 0 \Rightarrow x = -\frac{4}{3}\qquad x - 3 = 0 \Rightarrow x = 3 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{169}}{-6} \]
\[ x = -\frac{4}{3}\quad\text{or}\quad x = 3 \]

The parabola y = -3x² + 5x + 12

xy
-2.1667-12.917267
-1.16672.0829333
-0.166711.083133
0.833314.083333
1.833311.083533
2.83332.0837333
3.8333-12.916067

Questions

What are the solutions of -3x² + 5x + 12 = 0?

The solutions of -3x² + 5x + 12 = 0 are x = -4/3 or x = 3.

Does -3x² + 5x + 12 = 0 factor?

Yes: -1(3x+4)(x-3) = 0.

What is the vertex of y = -3x² + 5x + 12?

(5/6, 169/12). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 5/3, product = c/a = -4.

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