Question

Use quadratic equarion to solve
2x^2-7x-3=0

Answer 1

So whats the answer to x=

Answer 2

I dont need yucky decimals I need one good concerte answer

Answer 3

To solve the quadratic equation \(2x^2 - 7x - 3 = 0\), we can use the quadratic formula:

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

In this equation, \(a = 2\), \(b = -7\), and \(c = -3\).

First, we calculate the discriminant \(b^2 - 4ac\):

\[
b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot (-3)
\]
\[
= 49 + 24
\]
\[
= 73
\]

Now, we can substitute \(a\), \(b\), and the discriminant into the quadratic formula:

\[
x = \frac{-(-7) \pm \sqrt{73}}{2 \cdot 2}
\]
\[
= \frac{7 \pm \sqrt{73}}{4}
\]

So, the two solutions for \(x\) are:

\[
x_1 = \frac{7 + \sqrt{73}}{4}
\]
\[
x_2 = \frac{7 - \sqrt{73}}{4}
\]

These are the solutions to the equation \(2x^2 - 7x - 3 = 0\).

Answer 4

The solutions to the equation \(2x^2 - 7x - 3 = 0\) are:

\[
x_1 = \frac{7 + \sqrt{73}}{4}
\]

\[
x_2 = \frac{7 - \sqrt{73}}{4}
\]

To give numerical approximations:

1. **Calculating \(x_1\)**:

\[
x_1 \approx \frac{7 + 8.544}{4} \approx \frac{15.544}{4} \approx 3.886
\]

2. **Calculating \(x_2\)**:

\[
x_2 \approx \frac{7 - 8.544}{4} \approx \frac{-1.544}{4} \approx -0.386
\]

Thus, the approximate solutions are:

\[
x_1 \approx 3.886
\]

\[
x_2 \approx -0.386
\]

You can use these values as the solutions for \(x\).

Answer 5

Got it! The solutions to the equation \(2x^2 - 7x - 3 = 0\) in their exact form are:

\[
x = \frac{7 + \sqrt{73}}{4} \quad \text{and} \quad x = \frac{7 - \sqrt{73}}{4}
\]

If you prefer a single concise answer, you can indicate that the equation has two solutions:

\[
x = \frac{7 \pm \sqrt{73}}{4}
\]

This represents both solutions without using decimals.