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Para resolver a equação quadrática 2x² − 5x − 3 = 0, usaremos a fórmula de Bhaskara, que é x = -b +/- sqrt(b² - 4ac)2a.

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Para resolver a equação quadrática 2x2−5x−3=02x^2 - 5x - 3 = 0, usaremos a fórmula de Bhaskara, que é x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

Na equação dada, temos:
a=2a = 2
b=−5b = -5
c=−3c = -3

  1. Calcular o discriminante (Δ\Delta).

    Δ=b2−4ac\Delta = b^2 - 4ac Δ=(−5)2−4(2)(−3)\Delta = (-5)^2 - 4(2)(-3) Δ=25−(−24)\Delta = 25 - (-24) Δ=25+24\Delta = 25 + 24 Δ=49\Delta = 49
  2. Aplicar a fórmula de Bhaskara para encontrar os valores de xx.

    x=−b±Δ2ax = \frac{-b \pm \sqrt{\Delta}}{2a} x=−(−5)±492(2)x = \frac{-(-5) \pm \sqrt{49}}{2(2)} x=5±74x = \frac{5 \pm 7}{4}
  3. Encontrar as duas soluções para xx.

    Para x1x_1:

    x1=5+74x_1 = \frac{5 + 7}{4} x1=124x_1 = \frac{12}{4} x1=3x_1 = 3

    Para x2x_2:

    x2=5−74x_2 = \frac{5 - 7}{4} x2=−24x_2 = \frac{-2}{4} x2=−12x_2 = -\frac{1}{2}

    O conjunto solução é S={−12,3}S = \{-\frac{1}{2}, 3\}.

    Comparando com as opções fornecidas:
    A. S={−12,3}S = \{-\frac{1}{2}, 3\}
    B. S={12,−3}S = \{\frac{1}{2}, -3\}
    C. S={12,3}S = \{\frac{1}{2}, 3\}
    D. S={−12,−3}S = \{-\frac{1}{2}, -3\}
    E. S={1,−32}S = \{1, -\frac{3}{2}\}

    A opção correta é A.

    A. S={−12,3}\boxed{\text{A. } S = \{-\frac{1}{2}, 3\}}

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