Determine the most appropriate method and solve the following equation.
$2x^2-3x-5=0$

• A. Factoring; $x=2.142,x=-0.642$
• B. Square-Root Property; $x=\(\frac\)52,x=1$
• C. Quadratic Formula; $x=\(\frac\)52,x=-1$
• D. Quadratic Formula; $x=1,x=-\(\frac\)52$

Answer: (C)

Verified step by step guidance

1. Identify the type of equation given: \$2x^2 - 3x - 5 = 0\( is a quadratic equation because it is in the form \)ax^2 + bx + c = 0\( where \)a = 2\(, \)b = -3\(, and \)c = -5$.
2. Check if the quadratic can be factored easily by looking for two numbers that multiply to \(a \times c = 2 \times (-5) = -10\) and add to \(b = -3\). If factoring is difficult or not obvious, consider using the quadratic formula.
3. Recall the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula will give the exact solutions for any quadratic equation.
4. Substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula: \(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}\).
5. Simplify inside the square root (the discriminant) and then simplify the entire expression step-by-step to find the two possible values of \(x\).