To solve the quadratic equation $1.4x^2 - 2x - 10 = 0$ using the quadratic formula, we follow these steps.

Step 1: Identify the coefficients

The general quadratic equation is in the form: $ax^2 + bx + c = 0$ From the equation $1.4x^2 - 2x - 10 = 0$, we have:

• $a = 1.4$
• $b = -2$
• $c = -10$

Step 2: Apply the quadratic formula

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substitute the values of $a$, $b$, and $c$ into the formula: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1.4)(-10)}}{2(1.4)}$ Simplifying: $x = \frac{2 \pm \sqrt{4 + 56}}{2.8}$ $x = \frac{2 \pm \sqrt{60}}{2.8}$

Step 3: Simplify the square root

Since $\sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$, we have: $x = \frac{2 \pm 2\sqrt{15}}{2.8}$ $x = \frac{2}{2.8} \pm \frac{2\sqrt{15}}{2.8}$ $x = \frac{1}{1.4} \pm \frac{\sqrt{15}}{1.4}$

Step 4: Express the final solution

Thus, the two solutions are: $x_1 = \frac{1}{1.4} + \frac{\sqrt{15}}{1.4}, \quad x_2 = \frac{1}{1.4} - \frac{\sqrt{15}}{1.4}$

In simplified form: $x_1 = 0.714 + 1.936i, \quad x_2 = 0.714 - 1.936i$

These are the solutions in imaginary complex form.

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Do you want any more details on the steps, or have any questions?

Here are 5 related questions:

1. How does the discriminant help determine the type of roots?
2. Can the quadratic formula be used for non-real roots?
3. What happens when the discriminant is negative?
4. How do imaginary numbers relate to the complex plane?
5. Can any quadratic equation have complex solutions?

Tip: When solving quadratics with complex roots, always check the discriminant. If it's negative, the roots will be imaginary!