Questions: Solve the equation using the quadratic formula. 2x^2=2x-9 The solution set is □ b (Type an exact answer, using radicals as needed. Express complex numbers in terms of i. Use a comma to separate answers as needed.)

Solution Steps

To solve the quadratic equation \(2x^2 = 2x - 9\) using the quadratic formula, we first need to rewrite it in the standard form \(ax^2 + bx + c = 0\). Then, we can identify the coefficients \(a\), \(b\), and \(c\) and apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

First, we rewrite the given equation \(2x^2 = 2x - 9\) in the standard quadratic form \(ax^2 + bx + c = 0\).

\[ 2x^2 - 2x + 9 = 0 \]

Identify the coefficients \(a\), \(b\), and \(c\) from the standard form equation \(2x^2 - 2x + 9 = 0\).

\[ a = 2, \quad b = -2, \quad c = 9 \]

Calculate the discriminant \(\Delta\) using the formula \(\Delta = b^2 - 4ac\).

\[ \Delta = (-2)^2 - 4 \cdot 2 \cdot 9 = 4 - 72 = -68 \]

Since the discriminant is negative, the solutions will be complex numbers. Use the quadratic formula \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\).

\[ x = \frac{-(-2) \pm \sqrt{-68}}{2 \cdot 2} = \frac{2 \pm \sqrt{-68}}{4} \]

Simplify the solutions by expressing \(\sqrt{-68}\) as \(i\sqrt{68}\).

\[ x = \frac{2 \pm i\sqrt{68}}{4} = \frac{2}{4} \pm \frac{i\sqrt{68}}{4} = 0.5 \pm \frac{i\sqrt{68}}{4} \]

Since \(\sqrt{68} \approx 8.2462\),

\[ x = 0.5 \pm \frac{8.2462i}{4} = 0.5 \pm 2.0616i \]

Final Answer