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Determine the zero(s) of the quadratic function y=6x^2+13x+5

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write it small please

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To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \( a = 6 \), \( b = 13 \), and \( c = 5 \).

Step 1: Calculate the discriminant \( D \)

\[ D = b^2 - 4ac \] \[ D = 13^2 - 4 \cdot 6 \cdot 5 \] \[ D = 169 - 120 \] \[ D = 49 \]

Step 2: Apply the quadratic formula

Since \( D = 49 \) (which is positive), there will be two real roots.

\[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \] \[ x = \frac{-13 \pm 7}{12} \]

Step 3: Calculate the two solutions

  1. First root: \[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]

  2. Second root: \[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]

Final solution:

The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:

\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]

shorter please

To find the zeros of \( y = 6x^2 + 13x + 5 \):

  1. Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 6 \), \( b = 13 \), \( c = 5 \).

  2. Calculate discriminant: \[ D = 13^2 - 4(6)(5) = 169 - 120 = 49 \]

  3. Solve for \( x \): \[ x = \frac{-13 \pm 7}{12} \]

    • First root: \( x_1 = \frac{-6}{12} = -\frac{1}{2} \)
    • Second root: \( x_2 = \frac{-20}{12} = -\frac{5}{3} \)

Zeros: \[ x = -\frac{1}{2}, \quad x = -\frac{5}{3} \]

shorter please

To find the zeros of \( y = 6x^2 + 13x + 5 \):

  1. Discriminant: \[ D = 13^2 - 4(6)(5) = 49 \]

  2. Quadratic formula: \[ x = \frac{-13 \pm 7}{12} \]

  3. Roots: \[ x_1 = -\frac{1}{2}, \quad x_2 = -\frac{5}{3} \]

Zeros: \( x = -\frac{1}{2}, -\frac{5}{3} \)