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

Start by rewriting the quadratic equation in standard form: $2x^2 - 3x + 3 = 0$.

Identify the coefficients $a$, $b$, and $c$ from the equation $ax^2 + bx + c = 0$. Here, $a = 2$, $b = -3$, and $c = 3$.

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

Calculate the discriminant $b^2 - 4ac$. Substitute $b = -3$, $a = 2$, and $c = 3$ to find $(-3)^2 - 4(2)(3)$.

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.

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