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

$x=$\frac$34+$\frac{i\sqrt{15}$}{4},x=$\frac$34-$\frac{i\sqrt{15}$}{4}$

$x=$\frac$34+$\frac{5i}{4}$,x=$\frac$34-$\frac{5i}{4}$$

$x=$\frac{3+\sqrt{15}$}{4},x=$\frac{3-\sqrt{15}$}{4}$

$x=3+i$\sqrt{15}$,x=3-i$\sqrt{15}$$

Verified step by step guidance

First, rewrite the given equation in standard quadratic form, which is ax^2 + bx + c = 0. The given equation is 2x^2 - 3x = -3, so we can rewrite it as 2x^2 - 3x + 3 = 0.

Identify the coefficients a, b, and c from the quadratic equation. Here, a = 2, b = -3, and c = 3.

Use the quadratic formula to find the solutions for x. The quadratic formula is x = $\frac{-b \pm \sqrt{b^2 - 4ac}$}{2a}. Substitute the values of a, b, and c into the formula.

Calculate the discriminant, which is the part under the square root in the quadratic formula: b^2 - 4ac. Substitute the values to find the discriminant: (-3)^2 - 4(2)(3).

Since the discriminant is negative, the solutions will be complex numbers. Proceed to calculate the solutions using the quadratic formula, keeping in mind that the square root of a negative number involves imaginary numbers.

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