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The problem involves solving a quadratic equation using the quadratic formula. Based on the student's work, the quadratic equation being solved is 3x² − 13x + 10 = 0.

Mathematics

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Answer

x1=103x_1 = \frac{10}{3}

x2=1x_2 = 1

Step-by-step solution

8 steps

The problem involves solving a quadratic equation using the quadratic formula. Based on the student's work, the quadratic equation being solved is 3x2−13x+10=03x^2 - 13x + 10 = 0.

  1. Identify the coefficients of the quadratic equation.

    For ax2+bx+c=0ax^2 + bx + c = 0, we have a=3a=3, b=−13b=-13, and c=10c=10.

  2. Apply the quadratic formula.

    The quadratic formula is given by:

    x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  3. Substitute the values of aa, bb, and cc into the formula.

    x=−(−13)±(−13)2−4(3)(10)2(3)x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(3)(10)}}{2(3)}

  4. Simplify the expression.

    x=13±169−1206x = \frac{13 \pm \sqrt{169 - 120}}{6}

  5. Calculate the value under the square root.

    x=13±496x = \frac{13 \pm \sqrt{49}}{6}

  6. Calculate the square root.

    x=13±76x = \frac{13 \pm 7}{6}

  7. Calculate the two possible solutions for xx.

    x1=13+76orx2=13−76x_1 = \frac{13 + 7}{6} \quad \text{or} \quad x_2 = \frac{13 - 7}{6}

  8. Simplify the solutions.

    x1=206orx2=66x_1 = \frac{20}{6} \quad \text{or} \quad x_2 = \frac{6}{6} x1=103orx2=1x_1 = \frac{10}{3} \quad \text{or} \quad x_2 = 1

    The final answers are:

    x1=103orx2=1x_1 = \boxed{\frac{10}{3}} \quad \text{or} \quad x_2 = \boxed{1}
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