Solve the given quadratic equation using the quadratic formula. $2x^2-3x=-3$

• A. $x=\frac34+\frac{i\sqrt{15}}{4},x=\frac34-\frac{i\sqrt{15}}{4}$
• B. $x=\frac34+\frac{5i}{4},x=\frac34-\frac{5i}{4}$
• C. $x=\frac{3+\sqrt{15}}{4},x=\frac{3-\sqrt{15}}{4}$
• D. $x=3+i\sqrt{15},x=3-i\sqrt{15}$

Answer: (A)

Verified step by step guidance

1. Start by rewriting the quadratic equation in standard form: \(2x^2 - 3x + 3 = 0\).
2. Identify the coefficients \(a\), \(b\), and \(c\) from the equation \(ax^2 + bx + c = 0\). Here, \(a = 2\), \(b = -3\), and \(c = 3\).
3. Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values of \(a\), \(b\), and \(c\) into the formula.
4. Calculate the discriminant \(b^2 - 4ac\). Substitute \(b = -3\), \(a = 2\), and \(c = 3\) to find \((-3)^2 - 4(2)(3)\).
5. Since the discriminant is negative, the solutions will involve imaginary numbers. Simplify the expression under the square root and solve for \(x\) using the quadratic formula.