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Identify the coefficients of the quadratic equation.

Mathematics

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Answer

x=5±1096x = \frac{5 \pm \sqrt{109}}{6}

Step-by-step solution

5 steps

  1. Identify the coefficients of the quadratic equation.

    The given equation is 3x2−5x−7=03x^2 - 5x - 7 = 0.
    Comparing this to the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0, we have:
    a=3a = 3
    b=−5b = -5
    c=−7c = -7

  2. Write down the quadratic formula.

    The quadratic formula is:

    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=−(−5)±(−5)2−4(3)(−7)2(3)x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-7)}}{2(3)}

  4. Simplify the expression.

    First, simplify the terms:

    x=5±25−(−84)6x = \frac{5 \pm \sqrt{25 - (-84)}}{6} x=5±25+846x = \frac{5 \pm \sqrt{25 + 84}}{6} x=5±1096x = \frac{5 \pm \sqrt{109}}{6}

  5. Write the two solutions for xx.

    The two solutions are:

    x1=5+1096x_1 = \frac{5 + \sqrt{109}}{6} x2=5−1096x_2 = \frac{5 - \sqrt{109}}{6}
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