The equation you provided is a quadratic equation:

$x^2 - 3x + 4 = 0$

To solve for $x$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $x^2 - 3x + 4 = 0$, the coefficients are:

• $a = 1$
• $b = -3$
• $c = 4$

Now, plug these values into the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(4)}}{2(1)}$ $x = \frac{3 \pm \sqrt{9 - 16}}{2}$ $x = \frac{3 \pm \sqrt{-7}}{2}$

Since the discriminant $(b^2 - 4ac = -7)$ is negative, the solutions will be complex numbers. We express the square root of a negative number using imaginary units (where $i = \sqrt{-1}$):

$x = \frac{3 \pm \sqrt{7}i}{2}$

Thus, the solutions are:

$x = \frac{3 + \sqrt{7}i}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{7}i}{2}$

These are the complex solutions to the quadratic equation.

Would you like further details or have any questions?

Related Questions:

1. What are the conditions for a quadratic equation to have real or complex roots?
2. How do we simplify complex numbers in general?
3. Can quadratic equations with complex roots be solved graphically?
4. How do imaginary numbers apply in engineering or physics?
5. How does completing the square method compare with using the quadratic formula?

Tip:

Always check the discriminant ($b^2 - 4ac$) before solving, as it tells you whether the solutions will be real or complex.