Question

Find the solutions of 〖2x〗^2+3x=5 using the quadratic formula.

Answer 1

To solve the equation \( 2x^2 + 3x - 5 = 0 \) using the quadratic formula, we first identify the coefficients \( a \), \( b \), and \( c \) from the general form of a quadratic equation \( ax^2 + bx + c = 0 \):

- \( a = 2 \)
- \( b = 3 \)
- \( c = -5 \)

The quadratic formula is given by:

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

Now, we need to compute the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49
\]

Since the discriminant is positive, we have two distinct real solutions. Now, we can calculate the solutions:

1. Calculate \( \sqrt{{b^2 - 4ac}} \):

\[
\sqrt{49} = 7
\]

2. Substitute the values into the quadratic formula:

\[
x = \frac{{-3 \pm 7}}{{2 \cdot 2}} = \frac{{-3 \pm 7}}{4}
\]

This gives us two solutions:

1. \( x_1 = \frac{{-3 + 7}}{4} = \frac{4}{4} = 1 \)

2. \( x_2 = \frac{{-3 - 7}}{4} = \frac{{-10}}{4} = -\frac{5}{2} \)

So, the solutions of the equation \( 2x^2 + 3x - 5 = 0 \) are:

\[
x = 1 \quad \text{and} \quad x = -\frac{5}{2}
\]

Answer 2

shorter

Answer 3

To solve the equation \( 2x^2 + 3x - 5 = 0 \) using the quadratic formula:

1. Identify coefficients: \( a = 2 \), \( b = 3 \), \( c = -5 \).

2. Use the quadratic formula:
\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}
\]

3. Calculate the discriminant:
\[
b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49
\]

4. Substitute into the formula:
\[
x = \frac{{-3 \pm 7}}{4}
\]

5. Solve for \( x \):
- \( x_1 = \frac{{-3 + 7}}{4} = 1 \)
- \( x_2 = \frac{{-3 - 7}}{4} = -\frac{5}{2} \)

**Solutions:** \( x = 1 \) and \( x = -\frac{5}{2} \).